Study Unit

Calculus, Part 3

What This Text Covers . • •

A Review of the Derivative -------------------------------------------------------------------------------------------­Here the basic rules for calculating derivatives will be reviewed_

Implicit Functions and Parametric Equations --------------,--------------------------------------------------- 2 In this section, you will Jearn to calculate the first and second derivatives of implicit

functions and the derivatives of parametric equations by use of the chain rule_

The Derivatives of Trigonometric Functions ------------------------------------------------------------------ 9 This section deals with the derivatives of the six trigonometric functions and the six in­

verse trigonometric functions_

The Derivative of Exponential and Logarithmic Functions -------------------------------------------- 26 Here you will first review the meaning of exponential functions and logarithmic func­

tions_ You will then Jearn to calculate the derivatives of exponential and logarithmic

functions_

Applications of the Derivative ---------------------------------------------------------------------------------------- 39 This section starts with a review of the meaning of polar coordinates, and then explains

the use of polar differentials_ As you will see, polar differentials may be used to calculate

angles to the tangent line of a curve, angular velocity and acceleration, the relations be­

tween the rectangular and angular components of velocity, the acceleration of a re­

volving body, and circular motion with constant angular acceleration_

The Derivatives of Hyperbolic Functions ---------------------------------------------------------------------- 59 After you study the definitions of the hyperbolic functions, you will Jearn to calculate

the derivatives of the hyperbolic functions and of their inverses_

Appendix ------------------------------------------------------------------------------------------------------------------------ 65 Here you will find the solutions to the self-tests given in the text_

CALCULUS: FUNCTION AND USE Part 3

A Review of the Derivative

1. Student Objectives

In Part 1 we began the study of the method of calculus by considering the de­rivatives of polynomials. In Part 2 we learned the reverse process, integration, and

many of its applications. In this text we will consider the derivatives of other func­

tions, such as trigonometric, exponential, and hyperbolic functions and we will also discuss applications of these derivatives to angular measure.

When you complete this text, you will be able to:

• Calculate the derivative of an implicit function. • Use the chain rule to calculate the derivative of a function defined by para­

metric equations. • Calculate the derivative of any trigonometric function, any exponential

function with base e, any natural logarithmic function, and any hyperbolic function.

• Calculate the derivative of any compounded function. • Calculate the derivative of a function that is the product of any two or more

functions. • Calculate the derivative of any implicit function. • Calculate the angle that a tangent line to a curve at a point makes with the

point's radius vector. • Calculate the angular acceleration of a line when its angular position is de­

fined as a function of time.

Before moving on to derivatives of transcendental functions, we will review the formulas for calculating derivatives that we studied in Part 1 of this series.

2. Common Formulas for Derivatives

There are six essential formulas for calculating derivatives that you have learned

so far. Here the six formulas are listed in both derivative and differential form. You should have these formulas memorized by now.

du dv

de !(~)= v--u-

dx dx -=0 (1) (5) dx v2

d du d (du) - (cu)=c- (2) dx (un) = nun-1 dx (6) dx dx

d du dv d(c) = 0 -(u+v)=-+- (3) (7)

dx dx dx

d dv du d(cu) = c du - (uv) = u - + v- (4) (8)

dx dx dx

d(u+v)=du+dv (9) a(~\= v_du---;:;-_u_d_v v} v2

(11)

d(uv) = v du + u dv (10) (12)

Read through the list of formulas. Then close your text and write out each formula on a clean piece of paper. Check the formulas you wrote with the formulas listed to be sure that you have memorized them correctly.

The following self-test is designed to check your ability to apply the preceding formulas for differentiation. To get the given answer some extra algebraic manipula­tion may be required for some of the problems. You are urged to test your strength in algebra as well as in differentiation.

Find: in each of the following:

1. y = bx 3

2. y = b(x + 2) 3

3. y = 16(x 3 + 2x) 4

4. y = (x + bt (x 2 - cr

Self-Test 1

5. y = (3x 2 + 2x - 3)2 + .J x 2 + 12

6. y = (x2- 1)2(x2 + 2)3

(3x + 1)2

7. y= (2x2- 1)3

8. y = x(x 2 + 4x- 1)512

Implicit Functions and Parametric Equations

3. Change of Variable in Differentiation

Let us use the formula for the derivative of a product to see what the result will be when we change the independent variable. Equation 4, Art. 2, was given as

Letting u = x and v = y, then

dx since-= 1

dx

d du dv - (uv) = v- + u­dx dx dx

d dx dy - (xy)=y-+x­dx dx dx

dy =y+x-

dx (13)

If we needed to find the derivative of the same product (x y) with respect toy, then

. dy smce- = 1

dy

d dx dy -(xy)=y-+x­dy dy dy

dx =y- +x

dy

2

( 14)

and

Let us repeat the steps, but with a different product, such asx2 y 3 • Thus,

!!.cx2y3) = y3(2x) dx + x2(3y2) dy dx dx dx

= 2xy3 + 3x2 y2 dy dx

(15)

(16)

It should be both informative and interesting to note that equations 13 and 14 are

identical in differential form, which is

d(xy) = y dx + x dy (17)

A similar situation exists with equations 15 and 16:

(18)

The differential form is very useful, especially when the anti-derivative is desired, because

dx and dy, which are important parts of the integral, appear in the equation.

4. Differentiation of Implicit Functions

In most of the equations you have studied so far, the dependent variable was

expressed as an explicit function of the independent variable. Often it is either not

convenient or not possible to express one variable as an explicit function of the

other. In such situations the usual rules for finding the derivative can be applied to

each side of the equation and the desired derivative found as an implicit function

of the variables involved. Thus the equation may be differentiated as given. This

differentiation will result in another equation, which may be solved for the desired

derivative. This method can best be illustrated by the following examples:

Example 1

Given the equation of a circle x 2 + y 2 = 16, find the derivative dx· Solution

Since we are to find ;ix, we differentiate both sides of the equation with respect to x; thus

or

Therefore,

! (x2 + y2)= !(16)

dy 2x + 2y dx = 0

dy =-~ dx y

J

Ans.

Example 2

Find the second derivative,::~. if x 2 + y 2 = 16.

Solution dy X

From example 1 we know that -d = - -. Therefore, the second derivative is X y

dy X But-=·--· thus

dx y' '

Example 3

:x(d;) =~~=- !~) dy

y-x­dx =-yr-

Ans.

dy dx Given the equation x3y 2 = 16. Find-;- and d-, each as a function of x andy.

ax y

Solution

Let us begin the solution by using differentials; thus,

d(x 3y 2) = d (16)

3x 2y 2 dx + 2x 3y dy = 0

Dividing by x 2y and transposing one·term, we get

Dividing both sides by 2x dx, we get

dx . dy For dy , mvert dx and get

Example 4

2xdy =- 3ydx

dy =- 3y dx 2x·

dx 2x dy =- 3y"

Ans.

Ans.

Find the equation of the tangent line to the curve x 5 - y 4 + x 3 - y 2 = 0 at the point (1, 1 ).

Solution

Using differentials on the given equation, we get Sx4 dx- 4y3 dy + 3x2 dx- 2y dy = 0

Factoring,

Dividing by dx,

- (4y 3 + 2y)dy =- (5x 4 + 3x2)dx

dy (5x 4 + 3x2) dx = (4y3 + 2y)

Given, when x = 1 andy = 1 : = *= T• the slope of the tangent line. Using the point-slope form for

the tangent line, we get

4

Simplifying, we get

3y = 4x - 1. Ans.

This is the desired equation for the tangent tine to the curve.

Self-Test 2

Find: from each of the following equations:

1. (x- y)y2 + x + y = 0 x 3 y a

4. -+-=-y3 X b

2. y2 = x + v' x2 + y2 5. (x+y)(x-y)2 =b2

3. (y + x)lf2 + (y _ x)lf2 =a 6. x 4 -4x2y 2 +y 3 =0

. dy d2y . . Fmd- and - 2 from each of the followmg equations:

dx dx

7. 2x + xy + 3y = 6

8. y =.JX+Y 9. Find the equation of the tangent line drawn to the circle whose equation isx2 - 4x

+ y 2 + 6y = 12 at the point (5, 1).

10. Find the equation of the tangent line drawn to the curve y 3 = 4x2 - yx 2 at the

point (- 2, 2).

5. Introduction to Parametric Equations

When a particle moves in a curved plane path rather than in a straight lirie, it is cus­

tomary to specify its position at any time t by means of equations which give both x and

y as functions of t. For example, the equations of the trajectory of a projectile shown in

Fig. 1 have the form x = V0xt andy= V0 yt- ~ct2 , where Vox and V0 y are the x andy

components of V0 , which is the initial velocity at time t = 0.

y

X

Fig. I A trajectory with an initial velocity V 0

5

More generally, when discussing equations giving x andy as functions of time, we

write them in the general form x = g(t) andy = h(t). These two equations would specify

(x,y) in terms oft. Such equations are called parametric equations, and t is called the

parameter. Often we can eliminate t from the equations and obtain an equation of the

form y = f(x), in whichy is a function of x. From this equation we can get the derivative

dy = f'(x), which is the slope of the curve y = f(x). At this point, a question should dx ·

dy dy dx present itself concerning how dx is related to - = h'(t) and - = g'(t). The answer

dt dt

to this question is contained in the next article on the chain rule for derivatives. But be­

fore getting into that, let us develop the equation y = f(x) from the two given equations . X

for the trajectory, x = V0xt andy= V0yt- 1;2ct2. If x = V0xt, then t =-,which, when Vox

substituted into the equation for y, gives

y = V0 yt- Y2ct2

gives

x x 2 y = Voy- -Y2 c-2 = j(x)

Vox Vox

and the derivative is

dy Voy x , -=- -c- =f(x) dx Vox V2 Ox

6. Chain Rule for Derivatives

We have applied the chain rule for calculating the derivatives of compound

functions. Now we shall formalize this rule so that we can use it to calculate deriva­

tives of parametric functions.

RULE: If y = f(x) is a differentiable function of x. and x = g(t) is a differentiable

function oft, then y = f[g(t)] = h(t), which is also a differentiable function oft, and its

derivative with respect to time ish '(t)= f'(x)g '(t), or, in other symbols, dy = (dy) ( dx)· dt dx dt

This is called the chain rule of differentiation because it involves a chain of steps.

Ut us review these steps for finding the derivative of y with respect to t when y = f(x)

and x = g(t), or in short form, y = f[g(t) ]. We can see that here y is a function of a func­

tion, which is why we need to take the following three steps to find the derivative of y

with respect to t.

First, differentiate the functiony = fl.x) with respect tox to get dy = /(x). dx

dx Second, differentiate the function x = g(t) with respect to t to get-= g'(t).

dt

Third, multiply these two derivatives to obtain the third, dy. An example should dt

clarify this.

6

Example 1 Given: y = x 3 - 3x2 + 4x- 8 = j{x) and x = t 2 + 2t = g(t). Use the chain rule to find:.

Solution dy = 3x2 - 6x + 4 = 3(t2 + 2t)2 - 6(t2 + 2t) + 4 dx

dx dt = 2t + 2 = 2(t + 1)

: = (:)(!)= [3(t2 + 2t)2 - 6(t2 + 2t)+4) 2(t+ 1). Ans.

We could have arrived at this same answer by another route. That is, we could have first substituted the value of x in terms oft into the equation for y, which would give

When we differentiate with respect tot, we get

dy = 3(t2 + 2t)2(2t + 2)- 3(2)(t2 + 2t) (2t + 2) + 4(2t + 2) dt

Factor out (2t + 2) and obtain

dy = [3(t2 + 2t)2 - 6(t2 + 2t) + 4] (2t + 2) dt

which is equivalent to the first answer we obtained.

Since, from the chain rule, we got dr = (:)(:). it is possible to get ( :) by

dividing by (dx), as long as dx =I= 0. Doing so, we get dt dt

dy (dy/dt) -=--dx (dx/dt)

This form is particularly useful when we deal with parametric equations, such as x = g(t) and y = h(t), because it enables us to find the derivative of y with respect to x directly from these equations, as is illustrated in the following two examples.

Example 2

In Art. 5 were given the parametric equations for a trajectory: x = Voxt andy = Voyt- ~ct2.

dy Let us find dx"

Solution dx dy dt = Vox and dt = V Oy - ct

dy (dy/dt) Voy- ct dx = (dx/dt) = ---v;;;-

Substituting fort=~ ax' we get

dy Voy x , - =- -c --=f (x). Ans. dx Vox v5x

This is the same answer we obtained by a different method in Art. 5.

7

Example 3 dy

Given: the parametric equationsx = 2t + 3,y = t 2 - t. Let us find dx"

Solution

then

x-3 or, in terms of x, since t = - 2 -,

dx dt = 2 and

dy -=2t-1 dt

dy dy /dt (2t - 1) dx = dx/dt = -2-

dy X- 4 dx=-2-. Ans.

7. General Expression of Chain Rule

The chain rule is frequently expressed by using letters other than the letter t tore­

present the parameter. The most common of these is the letter u. Generally, y is stated

as a function of u and u as a function of x, which, in symbolic form, is written as

Y =flu) and u = g(x)

From the rule we get

~= (du)(:) We have already had an illustration of this formula when we developed the derivative

with respect to x of

y =un

Taking a derivative of both sides of this equation with respect to u, we get

dy -=nun-1 du

Invoking the chain rule, we get

du dx

Self-Test 3

For each set of parametric equations, fmd :.

1. X = 3t2 + 2, y = t2

3. y = u6 , u = 2 + x 2

4. y = u2 + 3u + 8, u = 2x + 3

2 5 y = 2 V2 + - v = (3x + 2)213 . v2'

8

The Derivatives of Trigonometric Funtions

8. Trigonometric Formulas

The following formulas of trigonometry, which should be sufficient for the purposes

of this text, are collected here for convenient reference:

sin A tanA =-­

cosA

I secA=-­

cosA

I cotA =-­

tanA

I cscA =-­

sinA

sin2 A +cos2 A= I

sec2 A=I+tan2 A csc2 A = I + cot2 A

sin(A +B)= sin A cosB +cos A sin B

sin(A - B) = sin A cos B - cos A sin B

cos(A +B) = cos A cos B - sin A sin B

cos(A- B) = cos A cos B + sin A sin B

( ) tan A+ tanB tan A +B = ----­

I- tan A tanB

sin 2A = 2 sin A cos A

cos 2A = cos2 A- sin2 A

cos 2A = 2 cos2 A - I

cos 2A = I - 2 sin2 A

. A ~-cosA sm-= 2 2

cos:i=. ~ 2 v~

sin A + sin B = 2 sin (A 2+ B) cos (A; B)

sin A- sinE= 2 sin (A; B )cos (A 2+ 8 )

f.A +B) (A- B) cosA + cosB= 2 cos\-2 - cos - 2 -

(A+B) (B-A) cosA-cosB=2sin - 2- sin - 2-

9

9. Circular Measure or Radians

The circle C in' Fig. 2 has a radius of r units. The arc lengths, between lines OX and

OB, is equal to the product r8, where 8 is the central angle, in radians. Thus if s = r8, then

the angle 8 =!..,which defines the angle in circular measure. r

8

c s X

Fig. 2 For the circle C, Arc-length s = r9

In differential form, the equation s = r8 is written as ds = rd8, where ds is the ele­mental arc length, a() is the elemental angle, and r is a constant.

For emphasis, let us restate the definition of an angle in circular measure:

The circular measure of an angle is the quotient of the length of an arc of a circle

divided by its radius. The radii that subtend the arc are the sides of the angle. Note

that this angle is the central angle () shown in Fig. 2.

Again for emphasis, let us say that the unit of the angle in this measure is the radian.

One radian is the angle for which the arc length s equals the radius of the circle. Any

angle may be said to contain a certain number of radians. Note, however, that the quo­

s tient- is dimensionless. It is customary t.:> speak of the angle 8 without using the word

r 2 1T

"radian." Thus, we may speak of the angle 2, the angle 3' the angle 4' the angle 2x, and

so on. During our early school days we learned that the circumference of a circle is 2rr r,

where r is the radius of the circle. At that time little attention was given to the factor 2rr.

Now we must note that 2rr is the number of radians in 360°, rr is the number of radians in

1T 180°, 2 is the number of radians in 90°, and so on. Of course, we should all know that

rr = 3.1415 9 to five decimal places. Usually we are permitted to round it off to 3.1416,

or even to 3.14, where precision is not demanded. In all work involving calculus, all angles which occur are understood to be expressed

in radians. In fact, many of the calculus formulas would be false unless the angles in­

volved were so expressed. The student should carefully note this fact, even though the

reason for it is not yet apparent. With this new information about angles in radians, such a trigonometric equation as

y = sinx

10

should take on added meaning. If the unit of x is not specified in degrees, it must be

taken to be in radians. Also, y = sin x may be considered as defining a functional

relation between two quantities exactly as does a simpler equation, such as y = x 2 . For

we may assign any arbitrary value to x and determine the corresponding value of y. This

may be done either by direct computation or by means of a table of trigonometric func­

tions, in which case we must interpret the value of x as denoting so many radians, which

again may be converted to so many degrees. Since many tables have angles in degrees, it

may be necessary to convert the value of x from radians to degrees. This can be done 180°

readily by multiplying it by the factor--. Of course, knowing the trigonometric func-. n

. o o o o o nnn tlons of 0 , 30 , 45 , 60 , and 90 , and that these angles in radians are 0, -, -, -, and

6 4 3

n l' is invaluable.

sin (.6.x) 10. Limiting Value of _...o.....-....:.

.6.x

One of the reasons for expressing an angle in circular measure is that it makes it pos­

;ible to show that

sin .6.x Lim--= I

~x + o .6.x

which is a very useful relationship, as we shall see. There are a number of ways to prove

that this equation is true. We shall do it by actually calculating the value of sin .6.x. That .6.x

is, we will substitute values for .6.x and sin .6.x and calculate the quotient. The values and

computed quotients are given in Table 1.

Angle .6.x deg

20 10 5 4 3 2 1 %

TABLE 1

VALUES AND COMPUTED QUOTIENTS FOR SIN .6.x AND .6.x

Angle .6.x sin .6.x Radians

0.34907 0.34202 0.17453 0.17365 0.08727 0.08716 o .. o6981 0.06976 0.05236 0.05234 0.03491 0.03490 0.01745 0.01745 0.00873 0.00873

sin .6.x -:;sx-0.97980 0.99496 0.99874 0.99928 0.99962 0.99971 0.99982 1.00000

sin .6.x Observe that .6.x did not have to become very small before -- became very nearly

.6.x

unity. This demonstration clearly indicates that the desired limit is unity.

The table also shows that sin ( x) = x, in radians, approximately, when the value of x is small.

11

11 L. . . V 1 f [ 1 - cos Llx] . amating a ue o Llx

The limit which was established in the preceding article may be used to find

the limit of P -~; Llx J as Llx approaches zero. Squaring both sides of the trigono­

metric identity

we get

. A Y 1- cosA Sln - =

2 2

2 sin2 A ·= 1 - cos A 2

Letting angle A = Llx, we get

I - cos Llx = 2 sin2 Llx 2

Then, dividing both sides by Llx,

1- cos Llx

Llx

Llx 2 sin2 -

2

Llx

. Llx . Llx sm2

=---=sm---Llx 2 Llx

2 2

. Llx sm-

Now, we already know from Art. 10 that Lim 2

--=1 Llx

And we also know that

Therefore,

12. Derivative of sin u

t;.x --+o

2 2

Llx Lim sin-=0

2 t:;.;c --+o 2

1- cos Llx Lim --Ll-x--= 0 x 1, or 0

t:.x-+ 0

From Part 1 of this series, remember that if y = j{x), the derivative of the func-

tionj{x) can be defined as the limit

dy Lim Lly Lim f(x + Llx) - j{x) -= -= dx Llx - 0 Llx Llx - 0 Llx

12

Here,j{x) =sin x, so j{x + 4x) =sin (x + 4x). From the list of trigonometric functions,

we see that

sin (A + B) = sin A cos B + cos A sin B

If A= x and B= ~.this identity becomes

So

So

j{x + 4x) = sin (x + 4x) = sin x cos 4x + cos x sin 4x

j(x + 4x) - j(x) = sin x cos 4x + cos x sin 4x - sin x = sin x (cos 4x - 1) + cos x sin 4x

dy _ Lim j(x + 4x) - f(x) _ Lim sin x(cos Ax - 1) cos x sin 4x -- - +-----dx 4x-O 4x Ax-0 4x Ax

Lim cos Ax - 1 Lim sin 4x : Sin X + A _ COS X A

4x-O 4x L.U.-0 ~x

: 0 X Sin X + 1 X COS X

:COS X

From Art. 10, we know that Lim sin Ax = 1 and from Art. 11 we know that Ax-0 Ax

Lim cos 4x - 1 = Ax-0 Ax

Lim -( 1 - cos Ax) = -0 = 0. Using the chain rule, if y = sin u Ax-0 Ax

and u is a function of x, dy = cos u x du. This new formula for calculating derivatives dx dx

can be stated as a rule:

. dy du Rule: If y = sm u, - = cos u x -

dx dx (19)

Formula 19 is the fundamental formula for finding the derivatives ofthe other trig­

onometric functions. We shall demonstrate this by developing the formula for the de­

rivatives of two other trigonometric functions. Before doing so, let's show by an

example how formula 19 is used.

Example

d" Find Jx when y = 4 sin (2x + 6).

Solution du

One can readily see that here u = 2x + 6, and dx = 2.

13

Therefore dy dx = ( 4) (cos u) (2), or 8 cos (2x + 6). Ans.

13. Derivative of cos u with Respect to x

The first stipulation is that u is a function of x which can be differentiated. From

the study of trigonometry, we have learned that

cos A = sin (90° -A)

Letting A = u and 90° = .!!:: , 2

cos u = sin ( ~ - u) Hence, by formula 19, the derivative

But

Therefore

=cos(~-u)!(~-u)

! (~- u) =- : and cos ( ~- u) =sin u

d . du - COS U = - SID U­dx dx

(20)

Observe the similarity and difference in these formulas. The derivative of the sine

function, formula 19, has a cosine function as a factor, while the derivative of the

cosine function, formula 20, has a sine function as a factor, but it has a negative sign.

J:::xample 1

Find: when y = 4 cos3 (2x + 6).

Solution Because the cosine is raised to a power, we will use a variation of formula {6) in Art. 2. In the variation

d n 1(dw) dx (w ) = nw"- dx

Let w = cos (2x + 6) and u = 2x + 6. Then

du = 2 dx

dw . du -= -sm u­dx dx

= - 2 sin (2x + 6)

Because y = 4w3, we use the variation of formula (6). Thus,

dy = 4 x 3w3- '(dw) dx dx

= [12 cos2 (2x + 6)][-2 sin (2x + 6)]

= - 24 cos2 (2x + 6) sin (2x + 6) Ans.

Example 2

Find: when y = x3 cos 2x.

14

Solution You should recognize that the right-hand side of the given equation is a product and that thcrc-

d du dv fore dx (1111) = v dx + u dx .

du dl' Let u = x3; then dx = 3x2. Let v =cos 2x; thus dx =- 2 sin 2x.

Therefore,

: = 3x2 cos 2x- 2x3 sin 2x. Ans.

Example 3 dy .

Find dx wheny = cos3 2x sin 4x.

Solution Here again we meet a product, and again we will use the formula for the derivative of a product.

Letting u = cos3 2x,

du 2 . ( dx) dx = 3 cos 2x(- sm 2x) 2 dx

=- 6 cos2 2x sin 2x

Also, letting v =sin 4x,

dv ( dx) dx = cos 4x 4 dx

= 4 cos 4x

Substituting in the formula for the derivative of a product, we get

:=(sin 4x) (-6 cos2 2x sin 2x) + cos3 2x (4 cos 4x)

or

: = -6 cos2 2x sin 2x sin 4x + 4 cos3 2x cos 4x. Ans.

14. Derivative of tan u with Respect to x Again the first stipulation must be that u is a function of x that can be differenti­

ated. From the study of trigonometry, we learned that the tangent may be expressed as the quotient of the sine and the cosine of the same angle. Expressed in the form of an equation, this is

sin u tanu=-­cos u

Since the right-hand side of this equation is a quotient, we must use the formula for the

derivative of !:L. However, to avoid confusion, we will use ~ instead of _vu • v v

Letting z = sin u,

and letting v = cos u,

dz du - =cosu-dx dx

dv =-sin u du dx dx

15

Thus we have

Substituting for z and v, we get

!!_(sinu\= cosu (cosu~)- sinu(-sinu~) dx cos~} (cos u)2

(cos2 u + sin2 u) du - dx - cos2 u

But from cos2 u + sin2 u = 1 and sec u =co~ u' we get

Example 1 dy Find dx, when y = 3 tan~-

d du -tan u = sec2 u­dx dx

Solution x du 1 Here u = 3; thus dx= 3· Now using formula 21, we get

Example 2

dy 3 5 (X) Find- wheny =-tan - · dx' 5 3

Solution Note here that the tangent is raised to the fifth power. Thus

: =i (5) tan4 (~):tan(~)= 3 [tan4 (~)] [sec2 (~~+ Therefore

15. Formulas for Differentiation of Trigonometric Functions

(21)

The formulas for the differentiation of trigonometric functions, where u represents any function of x that can be differentiated, are grouped here. Note that formulas 19 to 21 have already been dealt with in detail in Arts. 12 to 14.

d . du dxsmu=cosudx

d . du -cos u=-smu­dx dx

.!£..tan u = sec2 u du dx dx

16

(19)

(20)

(21)

d du - cot u =- csc2 u -dx dx

Jxsecu =secu tanu~

:fx esc u =-esc u cot u ~

(22)

(23)

(24)

The proofs of the last three formulas, 22 to 24, are similar to the proofs already

given. They are shown here to allow you to practice calculating derivatives and

working with trigonometric identities.

Let y = cot u = - 1- = (tan u) -I. tan u

Let y = sec u = - 1- = (cos u)-1• cos u

dy du Then-=- (tan u)-2 x sec2 u-

dy du Then-=- (cos u)-2 x (-sin u)-

dx dx dx dx

cos2 u du =---x--x-

sin2 u cos2 u dx

du =---x-

sin2 u dx

du =- csc2 u­

dx

I . I Let y = esc u = -:-- = (sm u)- .

SID U

Then dy = - (sin u)-2 cos u du dx dx

cos u I du =---x--

sin u sin u dx

du = - cot u esc u -

dx

sin u du =--x---

COS U COS U dx

du =tan u sec u­

dx

Study each of the following examples and solve the problems in the self-test

to be sure that you can work with the derivatives of the trigonometric functions.

17

Example 1

Given y = esc 3x, find : .

Solution By formula 24,

dy dx = -3 esc 3x cot 3x. Ans.

Example 2

Giveny = 4 sec2 4x, find dx· Solution

By formula 23, dy d dx = 4 (2) sec 4x dx sec 4x

= 8 sec 4x (sec 4x tan 4x) (4)

= 32 sec 2 4x tan 4x. Ans.

Example 3

Given r =a cot 38, find:~.

Solution By formula 22,

dr = a(-csc2 38) (3) = -3a csc2 38. Ans. d8

Example4

Given x = a(8- sin 8), find dx. d8

Solution dx d8 = a(l - cos 8). Ans.

Example 5

Giveny =cos (4x2 + 3x- 8), find:.

Solution dy . 2 d 2 dx = - sm ( 4x + 3x - 8) dx ( 4x + 3x- 8)

=- (8x + 3) sin (4x 2 + 3x- 8). Ans.

[Note that it is customary to place the algebraic or numerical factor in front of the trigonometric

factor.]

Example 6 Given the cycloid whose parametric equations are x = a(8- sin 8) andy= a(l- cr>~ 8). Find the

slope of the tangent to this curve at three points: a) where 8 = Yz1T; b) where 8 = 1r; and c) where 8 = 0.

Solution

We know that the slope of a curve in the X· andy-coordinate system is:~. Since we are not

given y as a direct function of x, we shall find the differentials dy and dx separately first. Then we

shall divide dy by dx to get the slope. Thus, since x = a(8- sin 8),

dx =a (d8- cos 8 d8) = a(l- cos 8) dtJ

18

and also y =a (1- cos 8)

dy = a[O- (-sin 8)] d8 =a sin 8 d8

Therefore

dy a sine d8 dx =a (1- cos 8) d8

But in this equation a and d8 appear in both the numerator and the denominator and thus may be can­

celled out to give the simpler expression

dy _ sin8 dx - (1 - cos e)

which we will now use to find the slopes.

a) When 8 = V21T (radians), or 90°, sin 'lz1T = 1 and cos 'lz1T = 0. Hence

dxdy] = 1 ! 0 = 1. Ans. e=Y21r

b) When 8 = 7T(radians), or 180°, sin 7T= 0 and cos 7T= -1. Hence

::]e=1T = 1 _o(-1) =% = o. Ans.

c) When 8 = 0, sin 0 = 0 and cos 0 = 1. Hence

dy] O O h. h . . d . A dx e = 0 = 1 _ (+1) = 0 w 1c 1s m etermmate. ns.

Self-Test 4

Find the derivative with respect to the indicated independent variable for each of the

following:

1. y =cos 6x 2. x = 4 sin 2t

3. y = 3 tan 3x

4. y = sin 2x cos 3x

5. y = sin2 3x cos 2x

6.y = 3x2 - sin Sx 7. y = 112 sec 2x + lf6 tan3 2x

8. r = a(sin 28 + cos 28) 1-sin3 8

9. r=---­cos e

10. xy + cotxy = 0

11. Find the slope of the tangent to the curve y = sin x at the point a) where x = ~ 1T and

b) wherex = 2. 12. In a tangent galvanometer, the current! is indicated by the deflection 8 of the needle

as compared with the deflection 8a produced by a known current/a, in accordance with

~e il the expression/=Ia--. Noting thatla and 8a are constants, find-.

tan 8a d8

16. Inverse Trigonometric Functions

The equation x = siny

19

defines a relation between the quantities x and y which may be stated by saying either

that x is the sine of angle y, or that y is the angle whose sine is equal to x. When we wish

to use the latter form of expressing x = sin y, we write instead the equation

y = sin-1 x

Here the - 1 is not to be understood as a negative exponent but as part of a new symbol

sin-1 • To avoid any possible misunderstanding, y = sin-1 xis sometimes written y =arc

sinx. These three equations have exactly the same meaning:

x =sin y y = sin-1 x y =arc sin x

It is necessary to be able to work with either form without difficulty.

In x = sin y, y is considered the independent variable, while in y = sin- 1 x, x is con­

sidered the independent variable. This equation then defines a function of x which is

called the inverse sine of x. It will add to the clarity of your thinking if you read ''y =

sin-1 x" as ''y is the angle whose sine is x." Similarly, if x =cosy, th~n y = cos1 x; if x =tan y, theny = tan-1 x; and so on for

the other trigonometric functions. In any of these equations it should be noticed that y is not completely determined

when x is given, since there is an infinite number of angles with the same sine, cosine, or

F 1 . I Th "f 1 1T 51T 137T tangent. or examp e suppose y = sm- x. en, 1 x =- ,y = -,-,-, and so on.

2 6 6 6 This creates a certain amount of ambiguity in using inverse trigonometric functions.

We have met this same situation when we went from the equation x = y2 to the

equation y =±...;X. Here if xis given, there are two values ofy.

In order that sin-1 x may be a single-valued function, it is necessary to limit the

• 1T 1T

range and adopt the convention that only angles between -2 and + 2 are accepted.

Thus y = sin-1 x is the angle, in radians, between-~ and+~, whose sine is x.

Similarly, y = tan-1 x is the angle, in radians, between - i and + ~· whose tangent

isx. The same convention applies to csc-I x, but a slightly different type of defini­

tion is required for cos-1 x, sec-1 and cot-1 x. For y = cos-I x,y is the angle, in radians, between 0 and rr, whose cosine is x.

For y = cot-1 x, y is the angle, in radians, between 0 and rr, whose cotangent is x.

For y = sec-1 x,y is the angle, in radians, between 0 and rr, whose secant is x.

17. Derivative of sin-1 u with Respect to x.

If y = sin -I u, where u is a function of x, then sin y = u. Take the derivative of

both sides of the latter equation with respect to x and solve for dx as follows:

sin y = u

dy du cosy-=­

dx dx

20

dy du dx =cosy x dX

However, it is necessary to calculate cosy when sin y = u.

Fig. 3 Right Triangle with angle y = sin-! u

Remember that the sine of any angle A is defined in terms of a right triangle

in which one angle measures A. Draw a right triangle with one angle measuring y

as is done in Fig. 3. The sine function is the ratio of the leg opposite the angle to the

hypo~enuse of the triangle. Sin y = u, or~. so let u be the length of the leg opposite 1

angle y. Then the hypotenuse of the triangle will be 1 and, by the Pythagorean

theorem, the leg adjacent to angle y isJ 1 - u2• Then

leg adjacent ~ __ r;----'21 _ u2 cosy= v 1 - u

hypotenuse 1

This value of cos y can be used in the above equation.

dy 1 du 1 du dx = cosydx = ~dx

or d . 1 du

dx sm-•u = ~ dx (25)

Thus we have a general formula for the derivative of an inverse sine function. Example _ d

Giveny = sin-1 v'1=4x2, wherey is an acute angle, in radians, find Jx. Solution -- 2

In the given equation, u = yl - 4x2 and u 2 = 1 - 4x . Apply formula 25 to get

dy I du I - 4x dx- .J I- (1- 4Xl)dx = ,J4?.x ~-

- 2 =~ Ans.

18. Derivative of arc sin (:)

Let y = sin -I(;} Then, by formula 25,

dy I I 1 dx Ax;= .J a2- x2

1-2 a

If you want, you can demonstrate this derivative without using formula 25. The

procedure is as follows: . I X . X

y = sm- -, so sm y =-a a

21

Taking the derivative of both sides of the equation for sin y,

dy I dy I cosy-=-, so-=---

dx a dx a cosy

It is necessary to find cosy. Draw a right triangle with angle y as is done in Fig. 4.

Sin y is the ratio of the leg opposite angle y to the hypotenuse of the triangle. Sin

y =~. so let x be the leg opposite y and let a be the hypotenuse. By the Pythogorean a

theorem, the leg adjacent to y is J a2 - x 2• Then

and

adjacent leg J a2 - x 2

cosy= =----hypotenuse a

dy = ---=-~~=::=- Ans

dx a cos y J a2 - x2 - -J a2 - x2 · ax

a

X Fig. 4 Right triangle with angle y = sin -t-;;

19. Derivative of cof1 u with Respect tox

Let y = cos- 1 u where u is a differentiable function of x. Just as we did in Art. 17,

let us convert the inverse function to cos y = u. The relationship between y and u is

shown by the right triangle in Fig. 5. Note that the side u is adjacent to the angle y here,

y

" Fig. 5 Right Triangle with angle y =cos-t u

though not in Fig. 3. Be careful about this difference. Taking the derivative of both sides

of the equation cosy = u, we get

. dv du - smy ::L. =-

dx dx

. . . dy 1 du DlVlde by- smy to get-=--.- -

dx smy dx

From Fig. 5, siny = v'}7; therefore,

d 1 du -cos-1 u =- -dx ~dx

(26)

22

Thus we have a general formula for the derivative of an inverse cosine function.

Ex=:: .. y = ,., -'(:'). Find : .

Solution 2 In the given equation, let u = :L. App;ying formula 26, we get

a

or

dy 1 d (x2) a 2x dx =- -/0----====2 dx \-;- =-vaL x4 -;;

dy 2x

dx=- vfa2-x4" Ans.

20. Derivative of tan-1 u with Respect to x

Let y = tan-1 u, where u is a differentiable function of x and the relationship be­

tween y and u is shown by the right triangle in Fig. 6. Again note that the side adjacent

to the angle y is given the value of unity.

If

then

and

or

u

Fig 6 Right Triangle with angle y = tan -1 u

dy du sec2 y-=­

dx dx

tany = u

and

But from Fig. 6, cosy=~ or cos2 y = 1 } u2

Therefore,

dy d ldu - =- tan-1 u = -- -dx dx 1 + uz dx

23

(27)

Example 1

. -l(x + 2) dy Gtven y = tan - 2- , f"md dx ·

Solution x+2 du 1

In the given equation, let u = - 2- ; hence dx = 2 .

Applying formula 27, we get

dy 1 (1) 4 (1) dx = 1 + (X; 2 r 2 = 4 + x 2 + 4x + 4 2

or dy 2

dx = x 2 + 4x + 8 · Ans.

Example 2

1 -1 ( b ) . dy Giveny =-tan -tanx fmd-ab a ' dx"

Solution

In the given equation, let u = (;tan x). Using formula 27,

dy 1 1 d (b ) dx = ab (b ) 2 dx ;tanx

I+ -tanx a

21. Formulas for Differentiation of Inverse Trigonometric Functions

The formulas for the differentiation of inverse trigonometric functions are listed

here. In each formula, u represents any function of x that can be differentiated. Note

that formulas 25 to 27 have been dealt with in detail in Arts. 17 to 20.

d I du - sin-1 u = -dx ~dx

d _1 I du -cos u=- -dx y1-U2 dx

d _1 I du -tan u=--­dx . I+ u2 dx

d _ I du -cot 1 u=-~­dx I+ u2 dx

24

(25)

(26)

(27)

(28)

Example 1

d -l 1 du - sec u = _ r;:;--; -dx uvu2- 1 dx

d _1 1 du -esc u=- -dx uyur.:-f dx

Giveny = cof1 yx2 + 2x, fmd :.

Solution

and

u =v'x2 + 2x

du 1 2x + 2 (x + 1) dx = 2 (x 2 + 2x)Y2 = yx2 + 2x

Applying formula 28, we get

dy 1 (x + 1) dx =- 1 + (x2 + 2x) y'x2 + 2x

dy (x + 1) 1 dx =- (x + 1)2 (x 2 + 2x)Y. =- -(x-+-1)-(x-,2=-+-2x-)=y..

Example 2

Giveny = csc-1( x; 2 ). find:.

Solution x-2 du 1

u =-2 - and dx =2

Applying formula 30,

Ans.

dy = 1 1 ---=2=-===-

dx ( x; 2) .(<x ~ 2)2 _ 1 2=- (x- 2)yx2- 4x

Example3

(29)

(30)

Ans.

_ 3x - 1 (3) dy dy dx G!Veny = (x 2 + 9) +cot -;,a) find dt; b) fmd the value of dt when x = 3 and dt = 9.

Solution The derivative of the quotient is

.!!_(~)- (x 2 + 9)(3)- 3x(2x)dx _ 27- 3x2 dx dt x 2 + 9 - (x 2 + 9)2 dt- (x 2 + 9)2 dt

The derivative of the inverse cotangtnt function is

! oof' (~ )·- 1 + ~~)'(~~): •- (<;'+ 9) (- ;.):

3 dx = x 2 + 9 dt

25

Combining the two terms, we get

dx Whenx = 3 and dt = 9,

dy [ 27 - 3x2 3 J dx dt= (x2 +9)2 +x2 +9 dt

27- 3x2 + 3(x2 + 9) dx (x2 + 9)2 dt

54 dx = (x2 + 9)2 dt. Ans. (a)

d~l 54 3 dt Jx=3 = (9 + 9)2 (9) = 2· Ans. (b)

Self-Test 5

Find: in each of the following:

1. y = sin-1 2x 2. y=tan- 1 (2x- 1) 3. y = cos-1 3x

d8 Find- in each of the following:

dx

7. 8 = sin- 1 ( 1 ) 8. 8 = ..!_ tan-1 ..!_ x

3 3

4. y = sec 1 3x 5. y = csc-1 (2x + 1) 6. y = coC 1 (x2 - 1)

9. 8 = tan-1 (x- 1) x+1

10. 8 = x sin- 1 x + .y'f+X2

The Derivative of Exponential and logarithmic Functions

22. What Is an Exponential Function?

An exponential function is one in which a variable appears in an exponent. Thus d",

52x, x2x, ex 2 , and alog x are exponential functions. Quantities such as x 5 and (x 3 + 2x)2

are NOT exponential functions, because the exponents are constants. They are

algebraic functions. We may do well to refresh our understanding of the properties of exponential func­

tions. Let's look at the equation y= bX

where b is any positive constant. If x = n, an integer, then y is determined by raising

b to the nth power by multiplication. If x =!!..., a positive fraction, then y is the qth q

root of the pth power of b, or the pth power of the qth root of b. In symbols, p - q

y = bq = ::;t;P And if X=- m y = b-m = - 1- .

' bm Furthermore, if x = 0, y = b0 = 1.

To illustrate these properties, let us take an equation such as y = bX and plot it on a

graph. We will assume that b = 1.5; then we will have the equation y = (1.5)X. A number

of positive and negative values have been assigned to x. The values were calculated and

tabulated as shown in Table 2, and plotted in Fig. 7. Observe that the curve in Fig. 7 is continuous for negative values of x as well as for

positive values of x, and that y is positive for all values of x. When x is a large positive

26

y

8 I 6 I 4 I

/ 2 ./

·" ·"'

·-· -·-· X

-4 -2 0 2 4 6

Fig. 7 Graph ofy = (l.S)x

TABLE 2

DATA FOR PLOTTINGy = (1.5)x

X y = (1.5)X

5 7.59375 4 5.0625 3 3.3750 2 2.2500 1 1.5000 % 1.2247 '4 1.1067 0 1.0000

-v.. 0.9036 -% 0.8166 -1 0.6667 -2 0.4444 -3 0.2963 -4 0.1975

value, y is a larger positive value. When x takes on large negative values, the curve ap­

proaches and is asymptotic to the line y = 0 or to the x-axis. When x is zero, y is

unity. When b is greater than unity, the graphs of all equations of the form y = bx

have a similar shape.

27

23. Logarithms

If a number x can be obtained by computing the result of placing an exponent

y on another number a, then y is said to be the logarithm of x to the base a. That is, if

X= aY

then y = loga x

These equations are simply two different ways of expressing the same fact concerning the

relation of x, y, and a. You should make yourself so familiar with these forms that you

can use one or the other as convenience may demand. From these equations, you can easily prove the fundamental properties of loga­

rithms. These are stated below in equation form as a reminder.

n IogaP = Ioga pn

Ioga I = 0

I log -=-log P a p a

Since you have had considerable experience with logarithms to the base I 0 [a= 1 0],

you should use this base to prove the preceding equations, and at the same time refresh

your knowledge of logarithms. When you are doing this, we suggest that you use easy

round numbers, such asP= I 000 and Q = I 00.

24. Base of Natural Logarithms

Theoretically, any positive number, except 1, may be used as the base of a sys­

tem of logarithms. Practically, however, there are only two numbers so used. The

first is the number 10, which is the base of the common system of logarithms. Com­

mon logarithms are very convenient for calculations and are used almost exclusively

with trigonometry. Another number, however, is more convenient in theoretical developments, since it

gives simpler formulas. It occurs often in problems in engineering and physics. This num­

ber is denoted by the letter e. Frequently, it is written as e, the small Greek epsilon. Un­

less a statement is made to the contrary, you are to understand that e (or e) is used to

denote the base of natural logarithms. These are sometimes referred to as Napierian or

hyperbolic logarithms. The value of e may be expressed by the infinite series

1 I I 1 1 e=I+-+-+-+-+-+

I 2! 3! 4! 5!

where 2! = 2 X 1, 3! = 3" X 2 X I, 4! = 4 X 3 X 2 X I, 5! = 5 X 4 X 3 X 2 X I, and so on.

Computing the infinite series to seven decimal places, we have

e = 2.71828I8 ...

28

For most practical purposes, however, it is sufficiently accurate to take

e = 2.7183

Since logarithms to the base e are called natural logarithms, we shall denote them by

the symbol Ln, wP,ere the letter L denotes a logarithm and the letter n denotes natural.

Thus the symbol Ln should be easy to remember. However, other texts may use the

symbol loge instead of Ln. The symbol log without any base indication will be taken to

mean a logarithm to the base 10. From the definition of a logarithm, we should know that if

X: eY

then y: Ln X

Caution and attention are very necessary when using tables and other texts. The

symbol for the natural logarithms may be written as log, and it will rest upon the reader

to know that the natural logarithm is meant. A good general rule to adopt is: Whenever

a logarithm appears in calculus, it is a natura/logarithm. A useful relationship that exists between logarithms of different bases is

Lnx log 10x =-- = log 10 e Ln x

Ln 10

from which we can gather that if .x = e, then

1 log 10 e=-­

Ln 10

The above relationships can be expressed more generally as

Lnb log a b = -- = log a e Ln b

Lna

Many calculators have a key for the natural logarithm and one for the common

logarithm of any positive number. Tables of natural logarithms exist and should

be used if available. For example, we can find from such a table that Ln 10 = 2.30259, or in a table of common logarithms log e = log 2.7183 = 0.43429, a:ad by simple

1 arithmetic, 2_30259 = 0.43429.

25. Derivative of ex

There are a number of ways to find the derivative of ex. Perhaps the simplest way is

to write ex as a power series and then to differentiate every term in the series.

In Art. 24, e was expressed as a series. Here it is obvious that e has the exponent I.

When the exponent of e isx, then the series is

x2 x3 x4 xs ex= 1 +x +- +- +- +- +

2! 3! 4! 5!

Taking the derivative with respect to x, we get

d 2x 3x2 4x3 5x4 -(ex)=O+ I+-+-+-+-+ ... dx 2! 3! 4! 5!

29

but

Hence

2 -= 1• 2! '

d x 2 x 3 x4 x 5 -(ex)= l+x+-+-+-+-+ ... dx 2! 3! 4! 5!

d Note that the right-hand sides of dx (ex) and ex are identical. Thus the derivative of

ex in series form is a series which is itself the series form of ex. That is,

which is unique! This is the only case in which the derivative of a function is equal to the

function itself! d

Note that since the derivative- (ex)= ex, the curve y =ex cannot have any turning dx

points-that is, no maxima or minima-because ex cannot be zero. The curve y =ex will

have a shape similar to the curve shown in Fig. 7.

26. Derivative of ebx

Let y = e bx, where b is a constant. Also, let z = ex; then z b = e bx, or y = z b. Now

we have y as a function of z and z as a function of x. Applying the chain rule gives us

dy dy dz -=--dx dz dx

Here dy - = bzb-1 dz

Also, since z = ex,

Hence

But

Thus dy -- = be bx e x ex = be bx dx

Therefore,

(31)

27. Derivative of eu

Let y = eu, where u is a differentiable function of x. Again using the chain rule, we

get dy = dy du

dx du dx

30

where

Therefore, d du -(e") = e"­d--: dx

This is a very useful formula.

Example 1

. - 3x2 dy G1veny = e , fmd ax·

Solution

Apply formula 32.

dy -3 2 d -3 2 -=e x -(-3x2)=-6xe x Ans. dx dx

Example 2

Giveny = e-3x sin 2x, find:.

Solution

dy -3x d -3x d - = e - (- 3x) sin 2x + e cos 2x-(2x) dx dx dx

=- 3e- 3x sin 2x + 2e- 3x cos 2x

= e-3x (2 cos 2x- 3 sin 2x). Ans.

Example 3

Giveny = sin- 1 (::: :=:). find :.

Solution d _1 1 du

Here we use dx sin u = y' 2 -, in which 1- u dx

and

4

31

(32)

Therefore,

Self-Test 6

Find : in each of the following:

1. y = e2fx 2. y = e-x2

5. y =x2 e3x

6. ex+ eY = 1 3. y = (2x2- 2x + 1)e2x 7. ex siny = 2

4. y=(3cosx+sinx)e 3x 8 sin x + cosx . y = ex

28. Derivative of Ln bx.

By the definition of a logarithm, if y = Ln bx, then bx = eY. Take the derivative

with respect toy and get

or

Hence

But y = Ln bx; therefore,

dx d b- =-(eY) =eY = bx

dy dy

dx -=x dy

dy =

dx X

dy d 1 - = - (Ln bx) = - = x- 1

dx dx X (33)

This result is very important in engineering and physics. It is also interesting to note that

it fills a gap that has existed until this moment. You will recall that we have derived and

d . 1 used the formula dx (xn) = n xn- , where n is any number, positive or negative, integral

or fractional. For example, we have found that

d dx (x2) = 2x

d - (x) = x 0 = 1 dx

32

and so on. From the preceding equations, observe that there is no power of x which,

when differentiated, will give an expression equal to x-1. In this article we have finally

found that this missing link is Ln x, whose derivative is 1/x or x-1•

29. Dtmvative of Ln u

Let y = Ln u, where u is a differentiable function of x. From the definition of a

logarithm we know that

u=eY

By the application of formula 32, the derivative with respect to x is

du d dy - =-(eY) =eY-dx dx dx

dy dy 1 du Solving for- we get-=--. Buty = Ln u and eY = u; therefore,

dx' dx eY dx

Example 1 3 dy

Given y = Ln 2x , find dx.

Solution

d 1 du - (Ln u) =-­dx. u dx

Here u = 2xl and we apply formula 34 to get

Example 2

Find:, givenxy =Ln (x + y).

Solution

(34)

Since this is an implicit function, take the derivative of both sides of the equation with respect

to x; thus

Transposing and factoring, we get

Solving for:.

dy 1 ( dy) x-+y=-- 1+-dx (x + y) dx

( 1 )dy 1 x- (x + y) dx = (x + y)- Y

dy 1- y (X+ y)

dx = x (x + y)- 1· Ans.

E_xampl~ 3 3 . dy . . . . . . . . G1ven y = Ln 2x , fmd dx' but s1mphfy the loganthm1c expressiOn before d1fferentJatmg.

33

Solution Here we write Ln 2x3 as Ln 2 + Ln x 3. Remembering that Ln x3 = 3Ln x,y = Ln 2x3 becomes

y = Ln 2 + 3Ln x and

: = 0 + 3(~ )= ~. Ans.

Because Ln 2 is a constant, its derivation is 0. Compare this result with the answer for example I.

Example4

Find:. given y = Ln (l- x 2)213.

Solution 2

Let u =(I - x2)3. Then

du 2 _.!.. d 2 --'--=-(1 -x2) 3-(1 -x2) =-(I -x2) 3 (-2x) = dx3 dx 3

and

Ans.

30. Derivative of au

Let y =au. It is understood that a is a constant and u is a differentiable function of x. Taking the natural logarithm of both sides, we get

Ln y = Ln au = u Ln a

Differentiating with respect to x, we obtain

or

Therefore,

Example 1 2 dy

Given y = a3x , find dx .

Solution

1 dy du du - - = u 0 + Ln a - = Ln a -y d.x d.x d.x

dy du -=yLna­d.x d.x

d du - (a")=a" Ln a­dx dx

~ ~ 2 Let u = 3x 2, then dx = 6x. Then apply formula 35 to get dx = 6x a3x Ln a. Ans.

Example 2

Find dy ify = x 2 3x dx .

Solution

(35)

We should recognize that y is the product of two factors. The derivative of x 2 with respect to x is 2x, and the derivative of 3x with respect to xis 3x Ln 3. Thus

dy d d -= 3x_(x2)+x2-(3x) dx dx dx

= 2x 3x +x 2 3x Ln 3. Ans.

34

Etamp/1.'3

.. dy tan- 1 x f-md- when v = 10 · ·

dx ·

Solution -I du I

Here we should recognize that a= 10 and u =tan x. Furthermore,- = --2 . according to dx l + x

dl· I 0 tan -I x . formula 27. Therefore, applying formula 35 we get--"- = l.n I 0. Ans.

' - dx I + x 2

31. Derivative of Loga u Let y =log au, where a is any positive number except 1 and u is any differentia­

ble function of x. Since we have already developed a rule for the derivative of a natural logarithm, we should try to convert loga u to a natural logarithm. We do this by applying

Lnu the rule mentioned in Art. 24. Hence loga u = L- = loga e Ln u. Here Ln a and loga e na are constants; Ln u is the variable. Therefore, differentiating each of these expressions, we get

Hence,

Example

dy d 1 1 du 1 du -=-(log u)=---- =log e-­dx dx a Ln a u dx a u dx

..

~(loga u) =_I_ du = loga e du dx uLna dx u dx

dy Given y = log sin x 2, find dx.

(36)

Solution In this equation the base of the logarithms is 10 and u is sin x 2. Applying equation 36, we

get

By formula 19,

Therefore,

dy loge d 2 - = -- - (sin x ) dx sinx 2 dx

dy loge dx = ----2 (2x) cos x 2 = 2x (log e) cot x 2. Ans.

smx

or, in terms of natural logarithms,

dy 2x cotx 2

dx - Ln 10

32. Using Logarithms to Simplify Differentiation

Ans.

The natural logarithm may be used to considerable advantage in two types of ex­pressions to simplify the work of obtaining the derivatives. The first type is an expression in the form of products, quotients, or a combination of products and quotients. The sec­ond type is an expression in which a variable has a variable for an exponent. The follow­ing two examples should help to clarify what we mean.

35

Example 1 /ex- l)(x- 2)

Differentiate y = V (x _ 3) (x _ 4) .

Solution Take the natural logarithm of both sides and get

Ln y = 1/2 [Ln (x- 1) + Ln (x- 2)- Ln (x- 3)- Ln (x- 4) J

Take the derivative of both sides with respect to x.

~ ~ = + [x ~ 1 + X ~ 2 -X ~ 3 -X ~ 4] 1 [ -2 (2x 2 - lOx+ 11) ]

= 2 (x- l)(x- 2)(x- 3) (x- 4)

dy (2x 2 -lOx+ ll)y -=-dx (x-l)(x- 2)(x- 3)(x-4)

(2x 2 - lOx+ 11) l(x- 1) (x- 2) (x- l)(x- 2)(x- 3)(x- 4) (x- 3)(x- 4)

2x 2 -10x+11 Ans.

Example 2 x Differentiate y = xe .

Solution Taking the natural logarithm of both sides, we get

X

Ln y = Ln x e = ex Ln x

Now we have a product which is easily differentiated. Doing so, we obtain

-- = ex - +ex Ln x = ex - + Ln x 1 dy 1 (1 ) y dx X X

dy (1 ) -=yex - +Lnx dx x

or

33. Graph of y = Ln x

Since we will be dealing with many expressions involving the logarithmic function, we should analyze this important and simple function to learn some of its characteristics. Let us graph the curve for y = Ln x. To do so we select a number of values for x (remem­ber that x must be positive) in the equation y = Ln x, and find the corresponding values of y. Doing this will give us a number of pairs of values for x andy. These are shown in

Table 3. Study this table carefully. Since the slope of the curve y = Ln xis given by dy = dx

1 -, we have tabulated the slope at the points we selected. When these points are plotted X

and connected by a smooth curve, we have a graph such as shown in Fig. 8.

36

TABLE 3

DATA FOR PLOTTINGy =Ln x

X !4 !6 1 2 4 6 8 10

y -1.386 -0.693 0 +0.693 1.386 1.782 2.079 2.303

dy 1 1 1 1 4 2 1 -

dx 2 4 6 8 10

Observe from Fig. 8 that the slope is positive for the limited portion of the curve

shown. However, if we analyze the equation of the slope of the curve,

dy 1 -=-dx X

we readily conclude that for all positive values of x the slope, : , will be positive. Hence

the graph of y = Ln x must rise steadily from left to right. Also, since the derivative is

continuous, the function Ln x is itself continuous, and the curve has a continuously turn­

ing tangent. Let us look. at the second derivative. Its equation is

d2y 1 dx2 =- x 2

y

3

slope= m 2 ·~

Lna Ln 10

X

0 4 6 8 10

-I

-2

Fig. 8. Graph of Curve y = Ln x

37

Note that for any value of x, positive or negative, the second derivative is negative. As we

have learned, this means that the curve's slope is everywhere decreasing. This fact

substantiates our earlier statement that the curve has a continuously turning tangent.

Observe also that the curve passes through the point ( 1, 0) because Ln 1 = 0. At this

point its slope is + 1, and so the tangent line at this point makes an angle of 45° with the

x-axis. Similarly, the slope at x = 4 is lf4 , and the slope at x = 10 is lfto, and so on.

As a final observation, we note that as x increases without limit, so the value of

y = Ln x increases without limit. That is,

Lim (Ln x) = oo x~oo

1 On the other hand, as x approaches zero through positive values, -will approach positive

X

1 infinity, but Ln x, which equals- Ln -,approaches- 00• Therefore, the curve reaches to

X

negative infinity as x approaches zero from the positive end.

34. Formulas for Exponential and Logarithmic Functions

The formulas for the differentiation of exponential and logarithmic functions are ac­

cumulated here for ready reference. In each formula u represents any function of x that

can be differentiated.

d _ (ebx) = bebx dx

d ( u) _ u du - e -e -dx dx

d 1 - (Ln bx)=- = x- 1

dx X

d 1 du dx (Ln u) =-;; dx

d du -(au) =au Ln a -dx dx

d dx log.z u

log.z e du du· =---=----

u dx uLna dx

Self-Test 7

Find: for each of the following:

1. y=Ln(x2 + 2x)

2.y=(Lnx)3

4. y =Ln cosx

5. y = Ln (tanx + secx)

2 3. y =log- 6. y=Ln(x+vX2+9')

x

38

(31)

(32)

(33)

(34)

(35)

(36)

x2 7. y=Ln 1+x2

8. y =x2 Lnx2

X 9. e<x + Y) = Ln­

y

11. xy = Ln (x + y)

12. y=Ln(Lnx)

13. y = 3sec x

10. y =Ln .J}+X2- X tan- 1 X 14. y =xLn x

Applications of the Derivative

35. Polar Coordinates

When determining the position of a point in a plane, we have been using two dis­

tances, x andy. The x-distance is perpendicular to they-axis, while they-distance is per­

pendicular to the x-axis. Thus we have the commonly called rectangular coordinates.

There are many other methods for determining the position of a point in a plane. A

very widely used method uses a distance, called the radius vector, and a directional

angle, called the vectorial angle. The radius vector is measured from a fixed point called

the origin, or the pole. The direction of the radius vector is measured by the angle the

radius vector makes with a fixed line, called the initial line, or polar axis. The distance

from the pole and the angle from the polar axis constitutes a system of coordinates re­

ferred to as polar coordinates. Study Fig. 9, in which we have the point P at a distance r from the origin 0 at an

angle () measured from the polar axis OX. Here we see that r and 8 completely describe

the position of point P. They are called the polar coordinates of point P.

Fig. 9. Position of Point P Described by Polar Coordinates r and ()

The quantities r and 8 may be either positive or negative. By accepted convention,

the positive angle is measured in a counterclockwise direction, as shown in Fig. 9. Hence

a negative value of 8 is laid off in a clockwise direction. Positive values of r are measured

from 0 along the terminal line of 8. Negative values of r are measured from 0 along the backward extension of the terminal line.

Let us examine Fig. 10 to clarify the meaning of negative and positive polar co­

ordinates. Observe that the point P 1 has the coordinates 6 and 60°. Both the radius

39

~ (G ,Go•J

~ (-5,-30°)

X

~(5,-30°)

~ (-G ,GO•)

Fig. 10. Examples of Points with Negative and Positive Polar Coordinates.

vector and the directional angle are positive. The point P2 has the coordinates ( -6, 60°).

Here the radius vector is negative, and is therefore measured from the origin along the ex­

tension of line OP 1 until it reaches point P 2•

Point P3 has a negative angle and point P 4 has a negative angle and a negative radius

vector. The point P4 could also be described as a point at(5, 150°) andP2 as a point at

( 6, 240°). Furthermore, P2 could also be described as a point at ( 6, -120°). From these

few examples, it should be obvious that the same point may have more than one pair of

coordinates, which is somewhat different from the rectangular coordinate system. In

practice, it is usually convenient to restrict r and () to positive values.

An equation using polar coordinates may define the locus of a point, and may be

plotted by adhering to the meaning of polar coordinates. Thus the equation r = 4 defines

a circle of radius 4 and the center at the origin, because the locus of all points at a dis­

tance 4 from the origin and in a plane is a circle. This is very similar to what we have

already met in rectangular coordinates. For example, the equation y = 4 defines a

straight line four units from the x-axis. In a polar coordinate equation using () but not involving a trigonometric func­

tion of fJ, it should be understood that() is always expressed in radians. For example,

to graph r = 8, the length of the radius vector at each point will be the value of fJ in

radians. When () = 180°, r will be rr or 3.14 and when fJ = 360°, r will be 2 rr, or 6.28.

36. Graphs with Polar Coordinates.

When an equation is given in polar coordinates, the corresponding curve may be

plotted by giving the angle () convenient values, computing the corresponding values of

the radius vector r, plotting the resulting points, and drawing a curve through them.

Plotting in polar coordinates is easier if you use paper ruled as in Figs. 11 and 12.

This kind of paper is called polar coordinate paper. The angle fJ is determined from the

numbers at the ends of· the straight radial lines. The value of r is counted off on the con­

centric circles, either toward or away from the number which indicates fJ, according as r

is positive or negative.

40

Example 1

Fig. 11. Graph of a Cardioid Determined by Equation r = 1.5 + 1.5 cos 8

Plot the locus of points given by the equation r = 1.5 (1 +cos e).

Solution To plot this curve, we assign values toe, and compute the corresponding values of r with the aid

of a table of natural cosines or of a calculator. Plot those points of the curve whose coordinates are thus

determined. Proceeding in this manner, we see. that as the values assigned to 8 increase from 0° to 90°, cos 8

decreases from 1 to 0; hence r decreases from 3.0 to 1.5 and we draw the corresponding arc ABC as

shown in Fig. 11. As 8 increases from 90° to 180°, cos 8 decreases from 0 to -1, r decreases from 1.5

to 0, and we draw the arc CDO. As 8 increases from 180° to 360°, cos 8 increases from -1 to +1,

causing r to increase from 0 to 3 in a continuous manner. We draw the arc OEFGA. If any more

values are assigned to 8, the corresponding points will follow the curve already drawn. Hence the

curve in Fig. 11 is the complete curve of the given equation. This curve is called a cardioid.

If we had t2kcn negative values of 8, the curve would have been traced in the reverse direction.

Example 2

Plot the locus of points given by the equation r = ( ~) 8.

Solution When plotting this curve, remember to consider 8 in circular measure (radians). When 8 is zero,

r = 0. As 8 increases, r also increases, so that the curve winds an infinite number of times around the

origin while receding from it, as shown in Fig. 12. For negative values of 8, the curve spirals in the re­

verse direction, as indicated by the broken curve. This curve is called the spiral of Archimedes.

41

Fig. 12. Graph of a Curve Determined by

Equation r = (~ J 8

Plot the following curves:

1. r = 4 cos 38 8

2. r = 2 sin 3

37. Polar Equation of Straight Line

Self-Test 8

3. r = 2 + 3 cos 8

4. r = 4 sin 28

Before we leave the subject of polar coordinates, we should derive polar equations

for some common geometric figures. First, let us try to derive the equation for a straight

line. Let the straight line LM in Fig. 13 be drawn perpendicular to the polar axis OX

at A, where OA =a. Let P(r,8) be any point on the line LM. Then OP = r is the hy­

potenuse of the right triangle OPA. By trigonometry,

or

Written in another form, we get

OP cos 8 = OA

r cos 8 =a

a r=-­

cos 8

This is the equation for a straight line perpendicular to the polar axis.

42

L

P(r,())

r

X

0

M

Fig. 13. Locus of All Points on Line LM Is

r = _!!:___e, the Polar Equation for a Straight cos

Line

a Let us test r = -- for its validity as an equation of a straight line. If we let 8 in-

cos e crease from 0° to 90°, cos 8 decreases from 1 to 0, which will cause r to increase from a

to oo, tracing out the lineAL. As 8 increases from 90° to 180°, cos 8 decreases from 0 to

-1. Thus r is negative. As r numerically decreases from -oo to a, it traces out the line MA.

If any other values are assigned to 8, no new points will be found. Therefore, we have

a established that r = -- is the equation of a straight line. One should note that the part

cos e AM of the line may also be found by letting 8 vary from 0° to -90°.

Of course, we should realize that the equation of this line in rectangular coordinates

isx =a.

38. Polar Equation of Circle

In Art. 35 we mentioned that the equation

r= 4

is a polar equation of a circle having a radius of four units with its center at the pole, or

origin. This same circle in rectangular coordinates is defined by the equation

vx 2 + y 2 = 4

Let's take a circle whose center is not at the origin but is on the initial line, b units

from the origin, and which passes through the origin, such as shown in Fig. 14. Since the

radius of the circle is b units, then OA = 2b. Also, let P(r,O) be any point on the circle.

Line segment OP = r. The triangle OPA is inscribed in a semicircle. We know that

triangle 0 P A is a right triangle, and that

OP= OA cos 8 or

r = 2b cos e which is the polar equation of the circle shown in Fig. 14.

43

c

X

A

Fig. 14. Locus of All Points on Circle C, Described by r = 2b cos()

Let us trace the curve, using r = 2b cos (). As the angle () is increased from 0° to

90°, r decreases from 2b to 0, and the upper half of the circle is constructed. As() is in­

creased from 90° to 18.0°, r is negative and decreases from 0 to - 2b, and the lower half of

the circle is constructed. Note that a negative r at an angle in the second quadrant ac­

tually places the point in the fourth quadrant. If any more values are assigned to(), the

points located will be the same as those already located.

It should be interesting to compare the preceding equation with the corresponding

one in rectangular coordinates, which is x 2 + y 2 = 2bx

Note the simpler form of the polar equation compared to the rectangular equation.

39. Length of an arc In Part 2 of this series, when we needed to find the length of a curve, we de­

veloped the relationship ds2 = d.x2 + dy2

where ds is the differential arc length, and dx and t~r are the differentials in terms of

the rectangular coordinates. As you know from analytic geometry, you can transform rectangular coordi­

nates to polar coordinates by using the following equations:

x = r cos 0 and y = r sin ()

With polar coordinates, both r and 8 are variables, so you can differentiate

d.x = cos 8 dr- r sin 0 d8

and dy = sin 8 dr + r cos 8 d8

Squaring both sides of these equations and substituting into ds2 = d.x 2 + dy 2, we obtain

ds2 = dx2 + dy2 =(cos 8 dr- r sin 8 d8)2 +(sin 8 dr + r cos 8 d8)2

= ( cos2 8 + sin2 8) dr2 + ( -2r sin 8 cos 8 dr a8 + 2r sin 8 cos 8 dr d8) +(sin28 + cos20)

r 2 d 82

ds2 = dr2 + r2 d 82 (37)

Study the following example to see how the formula can be used.

Example Find the length of the curve described by the equation r = fi that falls in the first quadrant.

Solution In the first quadrant, () will be between 0 and}-. First calculate dr:

dr = 2() d()

44

· Then calculate ds

ds = .J d? + r2 dff

ds = .J (26 d6)2 + 64 d62

ds = e,J4+82 d6

To calculates, find the integral of both sides of the equation

Let u = 4 + ff. Then du = 26 d6

I .!. s ="2Ju2du

I ~ =- u2 + C

3

I 3 =-(4 + ff-)2 + c

3 8

When 6 = 0, s = 0, so C = -3

3

s = f 6 ...;-;::;:er d 6

I - 8 When 6 :!!:, s = -2 ( 16 + 1r2) 2--. or about 2.82 units Ans.

2 4 3

40. Angles of Tangent Line There is an important relationship that can be observed from Fig. 15. Because

an exterior angle of a triangle is equal to the sum of the nonadjacent interior angles, from triangle OA P,

ct>= e + 1/1 (38)

where ¢ is the angle between the polar axis and the tangent line, () is the angle be­tween the polar axis and the radius vector, and 1/J is the angle between the radius vector and the tangent line.

In rectangular coordinates, when we want to discuss the direction of a curve at a point, we use the angle ¢ from the positive x-axis to the tangent line. In polar co­ordinates, it is more convenient to use the angle 1/J, which is the angle from the ex­tension of the radius vector to the tangent line T.

0

I I I I I I I I

8

r = f{9)

- P(r, 0)

Fig. 15. Relationship of Angles 0, 0 + 0

45

X

dy From the rectangular coordinate system we know that tan 1> = dx· From the

triangle OBP in Fig. 15, we obtain tan(} = 2:. because x = OB andy= BP. To calcu­x

late a formula for tan 1/1, use the equation cf> = 0 + 1/J, or 1/1 = cf>- 0. Then use the

formula from trigonometry

tan (A _B)= tan A- tan B 1 +tan A tan B

tan cP - tan 0 tan 1/J = tan ( 1> - 0) = ..,.------

1 + tan cP tan 0

S. dy mce tan 1> = dx

sin 0 dr + r cos 0 d 0 sin 0 ------.---and tan 0 = --. cos 0 dr -r sm 0 dO cos 0

Then, sin 0 dr + r cos 0 d 0 sin 0

cos 0 dr - r sin 0 dO cos 0 tan 1/1 = . 0 . 0 d dO sm (sm r + r cos 0

1 +cos 0 (cos 0 dr - r sin 0 dO)

sin 0 cos 0 dr + r cos2 0 d8 - sin 0 cos 0 dr + r sin2 () dO

tan 1/1 = cos 2 0 dr - r sin () cos () d() + sin2(J dr + r sin 8 cos() d8

r (cos2 () + sin2 0) dO tan 1/1 = .

( cos2 0 + sm2 O) dr

r d() tan 1/1 =--

dr (39)

To calculate a workable formula for tan!f>, use the fact that!f> = () + 1/J and the formula

tan A+ tanB tan (A + B) = as follows:

1- tanA tanB

tan (J +tan -,JJ tan <1> = tan ( (J + -,JJ) = -------'--

1 - tan (J tan -,JJ

The following examples will illustrate how these formulas can be used.

Example 1 a) Find the angle -,JJ in terms of the angle (J for a cardioid having the equation r =a (1 - cos £J),

as shown in Fig. 16. Also. find the angle that the tangent line T makes with the horizontal line OX,

b) when (J equals 120° and c) when (J equals 150°.

Fig. 16. Cardiod Having EqU4tion r = a(l -cos £J)

46

X

Solution a) From the equation of the curve we can find dr =a sin 8 d8, which we can substitute in !lqua­

tion 39 and get ,,,_rd8 _a(1-cos8)d8

tan 'I'- dr - a sin 8 d8

(1 - cos 8) Ans. sin 8

b) When () = 120°, cos() =cos 120° =-cos 60° = -T· and sin ()=sin 120° = sin 60° = v;

So 1/1 = tan-1 .J3 = 60°

Then,

(+.!. I -cos() 2

tan 1/1 = = ,-:;3 sin() y _, -2-

So, when() = 120°, the tangent line is parallel to the polar axis.

c) When()= 150°, cos()=- .J3 and sin fJ =J... 2 2

tan 1/1 = I - cos () sin fJ

lli

I

2

2

So 1/1 = tan- 1 3.73205 = 75°

Then.

= 2 + v'3 = 3.73205

Ans.

<P = () + 1/1 = 150° + 75° = 225°

The angle between the polar axis and the tangent line could be expressed as 225°. However, it should

be expressed as the smallest angle between the polar axis and the tangent line. Thus,

<P = 225° - 180° = 45° Ans.

<P = 45° because the tangent line is not a directed line. To say that two lines intersect at an angle of 45° is

the same as saying that they intersect at an angle of 225°.

Example 2 Given the equation of a curve r = 3 sin 38, which is illustrated in Fig. 17, fmd a) the angle as a

function of 8 that the curve makes with its radius vector; b) the slope of the tangent when e = 135°;

c) the angles 8, where the tangent is horizontal.

Solution a) From the equation of the curve we find

dr = 9 cos 38 d 8

The angle between the radius vector and the curve is 1/J and tan 1/J = r ::. Substituting for dr, we

obtain 3 sin 38 d8

tan 1/J = 9 cos 38 d 8

Dividing out the common factors, we get

or

sin 38 1 tan 1/J=--- =-tan 38

3 cos 38 3

l/l=tan-1 (113tan38]. Ans.

b) The slope of the tangent line is equal to tan</>, and

tan8+tani/J tan</>=1-tan8tanl/l

When 8 = 135°, tan 8 = -1; then 38 = 405°, tan 405° = tan 45° = + 1. Thus tan 1/J = 113 tan 38 = 113.

47

90°

120° Go•

240• 3oo·

270°

Fig. 17 .. Rose of Three Leaves Described by Equation r = 3 sin 38

.-: · Substituting these values into the equation for tan 1/J we get

c) When the tangent line is horizontal, tan </J = 0; thus

tan f! +tan 1J; · =0

1- tan e tan "'

Ans.

Substituting If3 tan 3e for tan lj;, we obtain

tan e + lf3 tan 3 f} ----"'---- = 0 1 - lf3 tan e tan 3 f}

Now our job is to find the angle f! that will satisfy this equation. It should be obvious that this equation will be satisfied when we find a value of f! that will make the numerator equal to zero. Set­ting the numerator equal to zero, we obtain

tan f! + If3 tan 3 f! = 0

To solve for f! in this equation it will be necessary to convert tan 311 into terms of tan e. We can do this as follows:

But

tan 2e +tan e tan (3e) = tan (2e +e) = 1 - tan 2e tan e

2 tan e tan (2e) =tan (e +e)= 1- tan2 e

48

Thus we get

2 tan e 1- tan2 6 +tan 6

tan 3e= 2tane 1-1-tan2etane

3 tan e- tan3 e 1- 3 tan2 e

Substituting this into tan 6 + 113 tan 36 = 0, we have

1 [3 tan e- tan 3 6] 0 tane+- = 3 1-3tan2e

Multiplying by the denominator and collecting terms, we obtain

2 tan e- 10f3 tan3e = 0

Factoring out 2 tan 6, we get

Now we have a product of two factors equal to zero and the rest is simple algebra. Thus 2 tan 6 = 0;

also, 1- Sf3 tan2 e = 0. From the f"rrst equation and by observing the curve in Fig. 17, we can readily see that the slope

of the curve is truly zero at 6 = 0. However, we must investigate the other factor. Setting it equal to

zero, we have

1- Sf3 tan2 6 = 0

or

tan2 e = 3fs

or

tan 6 = ±...(ii; = ±0.7746

Taking the positive value for the first quadrant, we find in the tables that

(J = 37.8°

Note that the negative value of tan 6 gives the angle 142.2°. The tangent line is also horizontal

at this point. You can see that this angle is about 142.2° by drawing on the graph of Fig. 17 a hori­

zontal line tangent to the curve in the f"rrst and second quadrants and reading off the angle 6 at the

point of tangency. You can see from the graph that the tangent is also horizontal when 8 = 270°. However, we did not

obtain this value in our result because tan 8 was used in the calculations and tan 270° is not defined. Hence,

the answer for c) has four parts: 0°, 37.8°, 142.2°, and 270°. This example shows how much of a background in trigonometric equations is required to solve prob­

lems involving polar equations. It also demonstrates the importance of using a figure to check the work.

Example3

lf the angle 8 in example 2 is increasing at a rate of 2 radians per second, at what rate is the radius vector increasing when 8 = 45°?

Solution Since the example calls for a. time rate of change of r, we need a derivative of r with respect to

t . F 1 2 h 3 · 3 Th d · · · h · · dr de 1m e. rom ex amp e , we ave r = sm 6. e envat1ve w1t respect to trrne IS- = 9 cos 38-dt dt"

49

-12Th fh d8 ·· 2 When 8 = 45°, 38 = 135° and so cos 38 = - - 2-. e angle's rate o c ange, dt' IS g1ven as .

dr Since dr is negative, r is decreasing.

dr d8 -= 9 cos 38-dt dt

An<

= 9 (-~ x 2 = -9J2or -12.7 units per second 2

Self-Test 9 1. Given the curve whose equation is r =a sin 30, find a) the angle which the curve makes with the initial line when 0 = 60°; b) the value of r when 0 = 60°; c) how fast the radius vector is increasing when 0 = 60° if the angle 0 is increasing at a rate of 72 radian per second.

o 2. Given the curve described by the equation, r =-,find a) the angle that the tangent to

1f

the curve makes with the radius vector when 0 = Jhrr; b) how fast the radius vector is in­creasing if the angle 0 is increasing at a rate of 2rr radians per minute. 3. Given the curve that is described by the equation r = 1.5 (1 +cosO), find a) the angle that the tangent line makes with the radius vector when 0 = 30°; b) the angle that the tangent line makes with the initial line when 0 = 30°; c) how fast the radius vector is in­creasing when the angle 0 is increasing at a rate of 2.0 radians per second.

41. Angular Velocity

In Art. 9 we discussed circular measure and found that the length of the arcs, sub­tended by the central angle 0, of a circle having a radius r is expressed by the formula

s = rO

When the angle 0 is increased, the arc length will be proportionately increased also. If 0 is a function of time, so wills be a function of time. We may, therefore, take the derivative of s = rO with respect to time and get

ds dO -=r-dt dt

Of course, r is a constant because the figure is a circle.

Now, ds is called the linear speed or velocity v of a point on the rotating radius at a dt

distance r from the center. Velocity v is also the measure of the rate at which s is de­scribed.

The rate of change of the angle 0 with respect to time is called the angular velocity of the radius vector. Angular velocity is commonly denoted by the Greek letter w (omega). Thus, by definition, we have

dO w=-dt

Substituting this into the preceding equation gives

ds dO v = dt = r dt = rw

which is a very useful relation between linear velocity and angular velocity as applied to a wheel or a circular path.

50

If 8 is in radians and t is in seconds, then w is in radians per second: When it is de­sired to have the angular velocity expressed in revolutions per second, we simply divide w

by 21T, since one revolution is equivalent to 21T radians. A popular symbol for angular velocity in revolutions per second is n. Thus, by definition again, we have

or

Example

w = 21Tn

w n=-21T

A wheel with radius 7.5 revolves with an angular velocity of 20 revolutions per second. What is the a) angular velocity, in radians per second, and b) linear velocity of a point on the rim of this wheel?

Solution

a) Since one revolution is equivalent to 2 rr radians, 20 revolutions per second is equivalent to 40rr radians per second, the required angular velocity. Ans.

b) Applying v = rw, in which r = 7.5, gives v = 7.5 (40Jrr) = 300rrunits per second. Ans.

42. Angular Acceleration

As you just learned, for circular motion, linear velocity is the distance traveled by a point on the circle per unit of time, and angular velocity is the amount of in­crease in the central angle per unit of time.

Angular acceleration is related to angular velocity just as linear acceleration is re­lated to linear velocity. Specifically, acceleration is the derivative of velocity with re­spect to time. The symbol for angular acceleration is a (Greek letter alpha), and it is defined by the relation

dw d 2 8 a=-=--

dt dt 2

Since, for circular motion, v=rw

then the linear tangential acceleration

dv dw at=- =r- =ra

dt dt

This establishes a useful relation between tangential acceleration and angular acceleration, as applied to a wheel or to any circular motion.

Example _ A wheel with radius 12 revolves so that its angular velocity is given by the equation w = 16 + 612, where

1 is in seconds and w is in radians per second. Find a) the angular acceleration when 1 = 4 second~; b) the linear tangential acceleration of a point on the rim when 1 = 4 seconds; c) the linear velocity of a point

on the rim as a function of time; d) the number of revolutions the wheel will make between 1 = 0 and 1 = 4 seconds.

Solution dw d 2

a) Angular acceleration is a= dt = dt (16 + 6t ) = 12t

When 1 = 4, the angular acceleration o: = 48 radians per second per second. Ans. b) The tangential acceleration is a 1 = ra Since r = 12, then a 1 = 576 units per second per second. Ans. c) The linear velocity is v = rw, or since r = 12,

v = 192 + 721 2 units per sec. Ans.

51

dO d) By defmition, w = ar· This must be written in differential form so that we can integrate to

fmci 0. Thus we get dO= wdt, which becomes f dO= f wdt = f (16 + 6t2) dt

or 0=16t+2t3 +C

When t = 0, 0 = 0, and thus C = 0. When t = 4, 0 = 16(4) + 2(64) = 64 + 128 = 192 radians. But 192

radians is equivalent to 1;~, or 30.56 rev. Ans.

43. Rectangular Components of Velocity

Let's consider a particle that traverses a circle at a· uniform rate of n rev. per sec

(revolutions per second.) Let P(x,y) be the position of the particle given in rec­

tangular coordinates, Fig. 18. Let OH be the projection of OP on OX, and OJ be

y

X

Fig. 18. Rectangular Projections of Point P on Circle

the projection of OP on OY. Drawing the lines HP, /P, and OP gives the right tri­

angle from which we may write the following relationships.

x = OH=r cos 0

and y = 0/ =PH= r sin 0

where r is the radius of the circle. We are given as the angular velocity of OP n rev per sec or 2nn radians per sec. In

formula form this is written

from which we obtain

Integrating this, we get

dO w=-=2nn

dt

dO= 2nn dt

0 = 2nn t+ C

If we consider that when t = 0, the particle is on OX, then C = 0.

Thus 0 = 2nn t = wt

Therefore, x = r cos 0 = r cos 2n n t = r cos w t

52

and y =r sin 8 =r sin 21rn t= r sin wt

The component of velocity parallel to OX is

V=dx X dt

Now, since we have x as a function of8 and as a function of wt, let us use both of these

functions to gain experience and show that we can arrive at the same answer by using

either.

Thus

But

x = rcos 8

dx d8 - =r(-sin8)­dt dt

d8 -=w dt

or

or

and

x = r cos wt

dx . d dt = r(- sm wt) dt (wt)

d - (wt)=w dt

Substituting win both equations, we get

V dx . . 8 =- =-rw sm wt=-rw sm X dt

Since 8 = wt, either form is satisfactory. How~ver, the form with wt is more frequently

used. Note that Vx is negative in the first quadrant, which means that x is decreasing

when 8 is increasing.

or

The component of velocity parallel to OYis Vy = dy. Sincey = r sin wt, dt

dy d - = r(cos wt)- (wt) dt dt

Vy = rw cos wt = rw cos 8

Having found the rectangular components of velocity, Vx and Vy, we can find the

resultant velocity by the right-triangle relationship

V=..JVx2 + V/

Substituting the values previously found for Vx and Vy, we get

V= ..J(- rw sin wt)2 + (rw cos wt)2

= ..J(rw)2 (sin2 wt + cos2 wt)

But (sin2 wt + cos2 wt) = 1

Therefore, V=rw

which is exactly what we found in Art. 41.

53

Example . A flywheel 4 units in diameter is rotating at a speed of 10 rev. per sec (revolutions per second). At a

given instant, point P is on the rim and I unit above the level of the center of the wheel, as shown in Fig. 19.

Find a) the velocity of P; b) the x-component of the velocity of P; c) they-component of the velocity of P.

Fig. 19. Point P on Rotating Flywheel

Solution a) The angular velocity of a spinning radius of the wheel is

w= 27rn

Here n = 10 rev per sec. Thus

w = 2 7r ( 1 0) = 20 7r radians per sec From V= rw,

V = 2(2071") = 407r = 125.7 units per sec. Ans.

b) In Fig. 19, PH= 1 and it is perpendicular to OX. Then triangle OPH is a right triangle, from

which we may write

sinO=%

and obtain the angle

e = sin-1 (%) = 30°

Using Vx = -rw sin Wt = -rw sin 8, and substituting for r, w, and 8, we find the horizontal compo­

nent of velocity to be

Vx =- 2 (207r) sin 30°

= -201T units per sec

= -62.8 units per sec. Ans.

c) In a similar fashion, obtain the vertical or y-component of velocity.

Vy=rwcosO

= 2(2071") cos 30° = 40rr(f)units per sec.

= 108.8 units per sec. Ans.

Self-Test 10

1. A point on the rim of a flywheel with a radius of 5 is 4 units above the level ofthe center of the wheel and has a horizontal component of velocity of- 120 units per sec. Find a) the vertical component of velocity for the given conditions; b) the angular speed, in revolutions per second.

2. The coordinates of a particle moving in a circle are x = 10 cos 4t andy= 10 sin 4t.

Find a) the x-component of velocity; b) the y-component of velocity; c) the re­sultant velocity; d) the x-component of acceleration; e) they-component of accelera­tion;[) the resultant acceleration;g) the angular velocity.

54

44. Acceleration While at Constant Speed An important concept in physics is the rate at which the tangent line to a cir­

cularly revolving body changes. For example, a record on a phonograph may be revolving at a constant speed. A point on the edge of this record will be moving at the same constant speed. However, remember that velocity is directed speed. The direction at the point is defined as the direction of the line tangent to the circle (or the record) at that point. Thus, the velocity of the point is constantly changing in direction.

In an earlier text we defined acceleration as the second derivative of directed displacement with respect to time or the first derivativ~ of velocity with respect to tiine. In mathematical symbols this is written for the x-component as

d2x dVx a =-=--

x dt2 dt and for they-component as

d2y dVy ay = dt2 =dt

For circular motion, we found in Art. 43 that Vx = -rw sin wt and Vy = rw cos wt

Thus the x-component of acceleration of a body in circular motion is

a =!!_ lv) =-rw2 coswt X dt \ X

for constant angular speed. Similarly, they-component of acceleration is

ay = :t (v.v) = -rw2 sin wt

also for constant angular speed.

or

The resultant centrifugal acceleration is

ac=Vai+a~

ac = V(-rw2 COS wt)2 + (-rw 2 sin wt)2

= ~2)2 (cos2 wt + sin2 wt)

=-rw2

Centrifugal acceleration is the rate at which velocity increases in the direction away from a line through the center of the circle. Centripetal acceleration is the rate at which velocity increases in the opposite direction.

From the above calculations, the centrifugal acceleration of the body rotating at a constant speed is negative. This means that the acceleration is centripetal ac­celeration.

This relationship is very interesting indeed. We see that the acceleration is con­stant in magnitude because both the radius and the angular speed are constants. The negative sign is significant because it means that the acceleration is directed along the radius towards the center of the circle. This offers an explanation why a bal­ancing force away from the center must be applied on a revolving body to keep it going in a circular path.

45. Circular Motion with Constant Angular Acceleration

Many problems lead us to situations where the angular velocity is not constant but changing. In Art. 42 we learned that the angular acceleration is related to angular

55

velocity by the equation.

dw -=a dt

which may be written as

dw= adt

If a is a function of time, we must find this before integrating; however, if a is a con­

stant, we can readily integrate and get

w=at+C1

If, at timet= 0, w = k, then the angular velocity at any timet is

w =at+ k

From the definition of angular velocity, we have

or

dO w=-=at+k

dt

dO=(at+k)dt

which, when integrated, gives

at2 o =-+kt+C2

2

If we begin counting 0 when t = 0, then c2 = 0, and the equation for angular dis­

placement becomes

at2 o = -+ kt

2

The equations for the position of a particle on a circle may now be written as

x = r cos 8 = r cos(~ a t 2 + kt)

y = r sin 8 = r sin (~a t2 + kt)

From these we get the equation of the velocity components

dO Vx = r (-sin 8) dt = -r (at+ k) sin 8

= -rw sin 8

dO Vy =r(cos8)-=r(at+k) cosO

dt

=rw cos 8

We can see that the preceding equations for the velocity components are similar to those

of Art. 43. The exception is that here w is not a.constant but a function of time.

By taking the time derivative of Vx, we obtain the x-component of acceleration,

56

which is

d (dw) . d8 a =- V = -r - sm 8 - rw (cos 8)-" dt " dt dt

B dw d d8 hi h h b · d · th din · · ut -d =a an -d = w, w c w en su stitute m e prece g equation gives us t t .

ax = -r a sin 8 - rw2 cos 8

= -r a sin w t- rw2 cos w t

Similarly, we get the equation for they-component of acceleration; thus

ay = :r Vy = r a cos 8- rw2 sin 8

=ra cos w t- rw2 sin w t

Note that these equations have components of ra, the tangential acceleration, mentioned

in Art. 42, as well as rw2 the centrifugal acceleration mentioned in Art. 44.

Let us go a step further and fmd the resultant of the components. Thus from

aR =.Jai +a~

we get

and

a;= (ra)2 cos2 8 -2 (ra) (rw 2 ) sin 8 cos 8 + (rw2 )2 sin2 8

Adding these two equations we get

ai +a; = (ra)2 (sin2 8 + cos2 8) + (rw2 ) 2 (sin2 8 + cos2 8)

Again we recognize that (sin2 8 + cos2 8) = 1, and so we get

ai + a; = (r a)2 + (rw2 ) 2 .

and the resultant acceleration

aR = ../(ra)2 + (rw2 ) 2

which shows us that ther.e is a right-triangle relationship between the components, such

as shown in Fig. 20.

Fig. 20. Radial and Tangential Components of Acceleration

57

The component ra. is usually called the tangential component of acceleration. It is

possible for a. to be negative, and thus for the component ra. to be directed in the op­

posite direction.

Example

A particle moves on a circle with a diameter of 8, according to the laws= 13 + 2t2, where sis the dis­

tance measured along the arc of the circle from a fixed point on the path to the moving point, and t is the

time, in seconds. When t = 2 sec, determine the following: a) the linear velocity of the particle; b) the an­

gular velocity of the radius to the particle; c) the tangent.ial component of acceleration; d) the angular

acceleration of the radius to the particle; e) the magnitude of the resultant linear acceleration of the particle.

Solution a) Since s is given as a function of time, we may find

V = ds = 3t2 + 4t dt

Thus when t = 2 sec, V= 3(4) + 4(2) = 20 units per sec. Ans.

b) s· f . I . 8 d ds d 8 h I I . . V 20 5 d. mce, orc1rcuarmotwn,s=r ,an di=r dt=rw,t eanguarveoc1tylsW=;-=4= ra 1ans

per sec. Ans.

d2s c) The tangential component of acceleration at= ra.=2.

dt When t = 2, a1 = 6(2) + 4 = 16 units per sec2 (second squared).

ds 2 d2s Since- = 3t + 4t, 2 = 6t + 4.

dt dt

at 16 2 d) The angula: acceleration a.= --; = 4 = 4 radians per sec Ans.

Let us solve for the angular acceleration in another way.

v 3t2 + 4t

F . d fi . . dw rom 1ts e m1t1on, a.= dt and

w = -; = --4- , from which we get

and when t = 2,

dw 6t+4 -;u=-4-

dw 12+4 2 a.= -- = --- = 4 radians per sec

dt 4

This agrees with the original solution. e) The centrifugal acceleration

ac = -rw2 = -4(5)2 = -100 units per sec2

Thus the resultant of the two components has the value

aR = Y(16)2 + (-100)2 = V10,256 = 101.3 units per sec2. Ans.

Self-Test 11

1. A line rotates in a vertical plane according to the law 8 = 4t3 - 12t2 , where 8 gives the

angular position of the line measured in radians and t is the time, in seconds. If the

radius of the circle is 2 and t = 3 sec, determine the following: a) the angular

velocity; b) the linear velocity; c) the angular acceleration; d) the tangential accel­

eration. 2. The velocity of a particle moving along a circular path varies according to the law

V = 2t3 - 6, where Vis the magnitude of velocity, in feet per second, and t is the time,

in seconds. If the radius of the circle is 2 and t = 2 sec, find the following: a) the

58

angular velocity of the radius; b) the angular acceleration of the radius; c) the tan­gential component of acceleration; cl) the magnitude of the resultant acceleration.

The Derivatives of Hyperbolic Functions

46. Definitions of Hyperbolic Functions

Since hyperbolic functions are used in solving certain engineering problems, we

should know what they are and how to use them. The name "hyperbolic functions" is

given to certain combinations of the exponentials e" and e-u. These combinations are

~(e" + e-") and ~(e" - e-"). They occur with sufficient frequency to merit special

names. The first combination is called the "hyperbolic cosine of u" and is defined as

cosh u = lh(e" + e-")

Note that the letter "h" identifies the function as a hyperbolic function, distinguishing it

from the trigonometric function cos u. The second combination is called the "hyperbolic sine of u," and it is defined as

sinh u = lh(e"'- e·")

The main reason for having the names of the hyperbolic functions resemble the names of the trigonometric functions is that the hyperbolic functions have many properties that

are analogous to the trigonometric functions. For example, the trigonometric functions such as cos u and sin u are easily identified

with the point (x,y) on the unit circle having the equationx2 + y 2 = 1. This is why they

are often called circular functions. By properly defining u, we may take x = cos u andy =

sin u. When we substitute for x andy in x 2 + y 2 = 1, we get the well-known identity

cos2 u + sin2 u = 1

In a similar fashion, the hyperbolic function cosh u and sinh u may be identified

with the point (x,y), but on a "unit hyperbola,"

x2- y2 = 1

If we letx =cosh u andy= sinh u,

then

and y 2 = sinh2 u = 14(e"- e-")2 = 74(e 2"- 2+ e-2")

When we subtracty2 fromx 2 , we get

x2- y2 = 74(e2u + 2 + e-2u)- 14(e2u- 2 + e-2u) = 74(4) = 1

which proves one of the basic relationships of hyperbolic functions, namely, that

cosh2 u - sinh2 u = 1

Thus The hyperbolic tangent is defmed similarly to that of the trigonometric function.

sinh u e" - e-u tanhu=--=--­

cosh u e" + e-u

59

The remaining hyperbolic functions are defined in terms of cosh u and sinh u as follows:

cosh u eu + e-u cothu=-- =---

Some useful identities are

sinh u eu - e-u

1 2 sechu=--=--­

cosh u eu + e-u

1 2 cschu =-- =--­

sinh u eu - e-u

cosh u + sinh u = eu

cosh u- sinh u = e-u

1- tanh2 u = sech2 u

coth2 u- 1 = csch2 u

From the first two identities it should be apparent that any combination of the expo­

nentials eu and e-u can be replaced by a combination of sinh u and cosh u, and vice versa.

47. Discussion of Hyperbolic Functions

Hyperbolic functions are of such importance and utility that their numerical values

have been calculated and tabulated at least for sinh u, cosh u, and tanh u. The values of

the other functions can be readily expressed in terms of these, so it is unnecessary to tab­

ulate all of them. Some slide rules also have sinh u and cosh u scales, and the values of

these functions may be· read directly. Certain major differences between the hyperbolic and circular functions should be

noted. The first difference is that the circular functions are periodic while the hyperbolic

functions are not periodic. The second great difference is that the two functions differ

in the range of values that they can assume; for example:

sin x varies between - 1 and + 1

sinh x varies from- oo to +oo

cos x varies between - 1 and+ 1

cosh x varies between + 1 to +oo

tan x varies from- oo to +oo

tanh x varies from - 1 to + 1

We shall conclude this article with certain formulas which you will find very useful.

sinh (x + y) = sinh x cosh y + cosh x sinh y

cosh(x + y) = cosh x cosh y + sinh x sinh y

sinh 2x = 2 sinh x cosh x

cosh 2x = cosh2 x + sinh2 x

cosh 2x + 1 = 2 cosh 2 x

cosh2x- 1 = 2 sinh2 x

Thus from the preceding you can see that practically all of the circular trigonometric

identities have hyperbolic analogies.

60

48. Derivative of Hyperbolic Sine Function

Lety =sinh u, where u is a differentiable function of x. Then

dy = ..!!_ sinh u = -~J.!.. (eu - e-u)] dx dx dx [2

Now we know the derivatives of exponential functions. Thus

Hence

Therefore,

d du -(eu)=eu-dx dx

and d ( -u) _ -u du - e --e -dx dx

d1 _ 1 du du - -(eu- e u) = -(eu + e-u)- =cosh u­dx2 2 dx dx

d du -sinh u =cosh u­dx. dx

Add this to your list of equations.

49. Derivative of Hyperbolic Cosine Function

Lety =cosh u, where u is a differentiable function of x. Then

dy d d [1 - 1 dx = dx (cosh u) = dx 2 (eu + e u)J

_ 1 ( u -u) du -- e - e -2 dx

Therefore, d du

dx (cosh u) = sinh u dx

SO. Derivatives of Other Hyperbolic Functions

(40)

{41)

Because the other hyperbolic functions of u are defined in terms of sinh u and cosh u, the formulas for their derivatives can be determined from formulas 40 and 41 along with the rule for calculating the derivative of a quotient. The formulas for the derivatives of the remaining hyperbolic functions, are as follows:

d du dx (tanh u) = sech2 u dx (42)

d du - (coth u) =- csch2 u-dx dx

(43)

d du dx (sech u) =- sech u tanh u dx {44)

d du dx (csch u) =- csch u coth u dx (45)

Observe that aside from the pattern of algebraic signs, these formulas are the exact ana­logs of the formulas for the corresponding circular functions.

Example 1

Giveny = sinh(3x2), fmd :.

61

Solution 2 du

In this problem u = 3x ; thus dx = 6x. Hence

dy 2 2 dx =cosh (3x ) (6x) = 6x cosh (3x ). Ans.

Example 2 dy

Giveny =tanh (1 +x 2), find dx'

Solution

dy = sech 2 (1 +x 2)~(1 +x 2) dx dx

=2xsech2 (l+x2). Ans.

Example 3 2 dy

Given cosh 3y = tan (3x ), find dx.

Solution d d

dx (cosh 3y) = dx (tan 3x2)

d d sinh 3y dx (3y) = sec2 (3x 2) dx (3x 2)

dy sec2 (3x 2) (6x) 3- = _ __;_ _ __;__ dx sinh 3y

dy 2x sec2 (3x2) -= dx sinh 3y

Ans.

Self-Test 12

Find : in each of the following:

1. y =sinh 2x 5. y = csch ( 1) 2. y = cosh2 (tanx) 6. y = Ln (cosh 3x)

3. y = tanh 2x 3 7. y = Y,. sinh 2x- ~x

4. y = cosh2 5x- sinh2 5x 8. coth 2y = tan 3x

51. Derivative of Inverse Hyperbolic Sine Function

Since we discussed the derivatives of the hyperbolic functions, we should be curious

about their inverse functions. The derivatives of inverse hyperbolic functions are similar to those of the circular functions with some differences in signs.

Let us follow through the familiar steps for finding the derivative of an inverse hy­perbolic sine of u written as sinh - 1 u.

Let y = sinh - 1 u, where u is a differentiable function of x. Then, from the definition of an inverse function, we get

u = sinhy

which we can differentiate with respect to x and get

du dy - =coshy­dx dx

62

or dy du -=---dx coshy dx

Now from cosh2 y- sinh2 y = 1, the fundamental relationship of hyperbolic functions,

we have

But

Hence

or

Substituting this into

we get

Therefore,

cosh2 y = 1 + sinh2 y

sinhy = u or

coshy=~

dy du =---

dx coshy dx

dy 1 du

dx=~ dx

d 1 du -(sinh- 1 u)= -dx ~dx

(46)

d 1 du Compare this with- (sin- 1 u) = y'l7 (equation 25), and note the sign difference.

dx 1-u dx

52. Derivatives of Other Inverse Hyperbolic Functions

We may readily establish the derivatives of the other hyperbolic functions in a

manner similar to that used for establishing the derivative sinh- 1 u. These formulas are

listed here for ready reference.

d 1 du - (cosh- 1 u) = ---=== dx Vu2=1 dx

d 1 du - (tanh- 1 u) = -- -dx 1- u2 dx

(The absolute value of u must be less than 1.)

d 1 du - (coth- 1 u) = --­dx 1- u 2 dx

(The. absolute value of u must be greater than 1.)

!!:._ (sech- 1 u) =- 1 du dx uvl-u2 dx

d _ 1 du - ( csch 1 u) = - -dx u~dx

(47)

(48)

(49)

(50)

(51)

The chief merit of the preceding inverse hyperbolic functions lies in their usefulness

in integration, which we shall study in Part 4 of this series.

63

Example 1 _ dy Given y = sinh 1 (2x), find dx·

Solution

Apply formula 46. du

Let u = 2x, and thus dx = 2. Therefore,

dy- 2 dx- VI+ 4x2

Ans.

Example 2 -1 dy

Given y = coth (sec x), find dx·

Solution

Apply formula 49. du

Here u =(sec x); thus dx =sec x tan x. Hence,

dy 1 sec x tan x sec x dx = 1- sec2 x (sec x tan x) = - tan2 x - --ta_n_x

sec x cosx sin x =- esc x. Ans.

Self-Test 13

Find: in each of the following:

l. y = sinh- 1 (3x) 6. y = sinh- 1 O~x)

2. y = tanh- 1 (sinx) 7. y = cosh- 1 (~)

3. y=coth- 1 e) 4. y = sech- 1 (cosx) 9. sinh- 1 (3x) = sinh (3y)

5. y = csch- 1 (x 2 ) 10. y = 2 tanh- 1 (tan ~x)

64

1. Formula (6)

y=bx 3

dy = 3bx3- 1 = 3bx2 Ans. dx

3. Formula (6)

y = 16 (x 3 + 2x)4 = 16u4

dy =4 X 16u4 - 1 = 64u 3

du

u =x3 + 2x

du 2 -= 3x + 2 dx

Appendix

Solutions to Self-Tests

Self-Test 1

2. Formula (6)

y = b (x + 2)3 = bu 3

u=x + 2

~= 1 dx

..!!!._ = 3bu 2 X du = 3b(x + 2)2 Ans. dx dx

dy = 64u 3 X du = 64 (x 3 + 2x)3 (3x2 + 2) Ans. dx dx

4. Formula (4)

y= (x + b)n (x 2 -c)m =u X v

u = (x + b)n

du n-1 -=n(x +b) dx v = (;2 -c)m

dv 2 m-1 - = 2mx(x -c) dx

dy dv du n 2 m-1 2 m n-1 -=u-+v-=(x+b) X2mx(x -c) +(x -c) Xn(x+b) dx dx dx

= 2mx (x + b)n (x2 -cr-1 + n(x2 -c)m (x + b)n-t Ans.

65

5. Formula (3)

y = (3x2 + 2x - 3)2 + Jx 2 + 12 = u + v

u = (3x2 + 2x- 3)2

du 2 - =2(3x + 2x- 3)(6x + 2) dx

v = (x 2 + 12)112

dv 1 2 -112 x d- = -2 (x + 12) (2x) = _2 ___ 11_2

x (x + 12)

X

dy du dv 2 x -=- +-= 2(3x + 2x- 3)(6x + 2) + Ans. dxdxdx ~

6. Formula (4)

7.

y = (x 2 - 1)2 (x2 + 2)3 = u X v

u = (x2 - 1)2

du 2 - =4x(x -1) dx

v = (x 2 + 2) 3

dv 2 2 - = 6x (x + 2) dx

dy dv du 2 2 2 2 2 3 2 - =u- + v -= (x -1) X 6x(x + 2) + (x + 2) X 4x(x - 1) dx dx dx

= 6x(x2 - l)(x2 - 1)(x2 + 2)2 + 4x(x2 + 2)(x2 - 1)(x2 + 2)2

= [6x(x2 - 1) + 4x(x2 + 2il (x2 -1)(x2 + 2)2

= 2x(3x2 - 3 + 2x2 + 4)(x2 -1)(x2 + 2)2

= 2x(5x2 + 1)(x2 - l)(x2 + 2)2 Ans.

Formula (5)

(3x + 1)2 u y=

(2x2 - 1)3 v

u = (3x + 1)2

du dx = 6(3x + 1)

v = (2x2-1)3

dv = 12x(2x2 -1)2 dx

66

du dv dy v~-u~ (2x2 -1)3 X6(3x+l)-(3x+l)2 Xl2x(2x2 -1)2

v2 (2x2 _ l)6

6(2x2 - 1) (2x2 - 1)2 (3x + 1) -12x(3x + 1) (3x + 1) (2x2 - 1)2

(2x2 - 1)6

= [12x2 - 6- 36x2 -12x] (2x2 -1)2(3x + 1) = -6(4x2 + 2x + 1)(3x + 1) Ans

(2x2- 1)6 (2x2 -1)4 .

8. Formula (4)

y =x(x2 + 4x -1)512 =u X v

u=x

du - =1 dx

2 5/2 v= (x + 4x -1)

dv = ~ (x2 + 4x- 1)312 (2x + 4) = S(x2 + 4x -1)312 (x + 2) dx 2

dy = u dv + v du = Sx(x + 2) (x2 + 4x- 1)312 + (x2 + 4x- 1)512 X 1 dx dx dx

= (Sx2 +lOx) (x2 + 4x -1f12 + (x2 + 4x -1) (x2 + 4x -1)312

2 2 2 3/2 2 2 3/2 = (Sx + lOx + x + 4x -1) (x + 4x- 1) = (6x + 14x- 1) (x + 4x- 1) Ans.

Self-Test 2

1. (x-y)y2 +x+y=O

( ) d(y)2 2 d(x-y) x-y ~+y ~

dx dy +- +- =0

dx dx

(x-y)2y dy + 'i-dy)y2 + 1 + dy = 0 dx \' dx dx

dy 2 dy 2 2 dy dy 2xy ~ - 2Y dx + y - Y dx + 1 + dx = 0

dy 2 dy dy 2 2xy - - 3y - + - =- y - 1

dx dx dx

dy 2 2 dx (-3y + 2xy + 1)=-y -1

dy -y2 -1

dx = -3y2 + 2xy + 1 3y2 -2xy-l Ans.

67

+ 2y-dy) dx

dy

Ans.

2 2 112 2y(x + y ) -y

1/2 112 3. {y+x) +(y-x) =a

~(y + x)-112 (ay + 0 + ~(y -x)-112(dy _) = 0

2 \d.x J 2 dx 1}

dy 1 1 dy X-+--+ ---X---- =0

.JY+X dx ...;y+x .jy _ x dx ~

(Jy:x + Jy~x) ix = ~

(.JY-X+.JY+X) dy = JY+X-~ .../Y+X ~ dx JY+X ,jy -x

~-~ -= Ans. dy

dx a

dy x dx- y

------- + = 0 x2

68

dy xy4 y dx = x2y3 = ; Ans.

5. (x + y)(x-y)2 =b2

(1 +ix) (x-y)2 +2(x+y)(x-y) (~-:)=o 2 2dy . dy

(x-y) + (x-y) dx +2(x+y)(x-y)-2(x+y)(x-y)dx=O

2 dy 2 [(x-y) -2(x+y)(x-y)) -=-(x-y) -2(x+y)(x-y)

dx

dy

dx

-(x-y)- 2 (x + y)

(x-y)- 2.(x + y)

-x + y - 2x - 2y 3x + y = -- Ans.

X - y - 2x - 2y X + 3y

3 2 2 dy 2dy 4x - 8xy - 8x y - + 3y - = 0

dx dx

2 2 dy 2 3 (3y - Sx y) - = 8xy - 4x

dx

dy

dx

7. 2x + xy + 3y = 6

8xy2 -4x3

3y2 - Sx2y

dy dy 2+y+x- + 3dx=O

dx

dy dy X-+ 3-=-y-2

dx dx

dy -:Y- 2 Ans. dx = ---;-+3

Ans.

69

d2y (x + 3)(- fx)+ (y + 2) X 1

dx 2 (x + 3)2

(x + 3)(y + 2) + y + 2 x+3

8. y = (x + y)

1 2

(x + 3)2

dy = ~(X+ y)_l/2 ~ + dy ) dx 2 \ dx

dy 1 1 +

dx

X dy =

X dy dx

dy

dx dx 2.Jx + y

2y + 4

(x + 3)2 Ans.

dy

dx --- Ans.

2v'X'+Y- 1 2y - 1

~ -2 ~ 1 ~ - 2 =-(2y-l) X 2- =- X 2-dx dx (2y- 1l dx

2 x---(2y- 1)2 (2y- 1)

dy dy 2x - 4 + 2y - + 6- = 0

dx dx

dy (2y + 6)- = 4 - 2x

dx

dy 4- 2x

dx = 2y + 6

Forx=5,y=1

dy 4-10 -6 -3 m=-= --=-=-

dx 2+6 8 4

(y-yl)=m(x-xl)

-3 (y- 1) = 4 (x- 5)

4y -4=-3x + 15

3x + 4y = 19· Ans.

70

2 Ans.

2dy 2dy 3y - = 8x-(x - + 2xy)

dx dx

2 2 dy (3y + X ) dx = 8x - 2xy

dy 8x-2xy

dx 3y2 +x2

For x =-2,y = 2

dy -16 + 8 m = dx = 12 + 4 2

(y -yl) = m(x -x 1)

1 (y-2)=-2(x+2)

2y -4 =-x-2

dx - = 6t dt

y = t2

dy - = 2t dt

x + 2y=2 Ans.

3. y=u6

dy = 6us du

u= 2 +x2

1=2xdx du

dx -=-du 2x

dy s 1 s dx = 6u + 2x = 12u x Ans.

Self-Test 3

dx 2 - = 3t dt

dy - = 2t dt

dy 2t 2 -=-=-AnSa dx 3t2 3t

4. y = u2 + 3u + 8

71

dy -=2u+3 du

u=2x+3

1 = 2dx du

dy 1 dx = (2u + 3) + 2 = 4u + 6 Ans.

S. y = 2V2 + 2V-2

dy -3 . 4. 40 - 4 dV= 4V-4V =4V- vl = vl

V = (3x + 2)213

1 = ~(3x + 2)-113 (. :v) dx (3x + 2) 113 -= dV 2

dy =(40 -4\. (3x + 2)113

dx v3/ 2

8(0 -1)

vl(3x + 2)113 Ans.

2/3 6. y = w

dy 2 -1/3 -=- w

dw 3

1 = 2x(J:)

dy 2 - 113 . 1 4x dx = 3w ..,. 2x·= 3w1/3 Ans.

Self-Tesf 4

1. y = cos6x=cos u

u =6x

du =6 dx

dy 6"6xA - =- s1n ns. dx

3. y = 3 tan 3x = 3 tan u

u = 3x

du = 3 dx

dy = 9 sec2 3x Ans. dx

S. y=sin2 3xcos2x=uXv

u = sin2 3x

du = 6 sin 3x cos 3x dx

2. x =4sin 2t=4 sinu

u = 2t

du =2 dt

dx - = 8 cos 2t Ans. dt

4. y = sin 2x cos 3x = u X v

72

u=sin2x

du -=2cos2x dx

v=cos3x

dv . - =-3sm3x dx

dy = 2 cos 2x cos 3x - 3 sin 2x sin 3~ Ans. dx

v =cos2x

dv . - =-2sm2x dx

dy = 6 sin 3x cos 3x cos 2x - 2 ·sin 2x sin2 3x Ans. dx

6. y = 3x2 - sin Sx

7.

dy - = 6x - S cos Sx Ans. dx

1 1 3 1 1 y=-sec2x+-tan 2x=-u+-v

2 6 2 6

u=sec2x

du -=2sec2xtan2x dx

v = tan3 2x

dv 2 2 -=6tan 2xsec 2x dx

dy 1 du 1 dv -=- -+--dx 2 dx 6 dx

= sec 2x tan 2x + tan2 2x sec2 2x Ans.

9. r 1-sin3 9

cos 9

u v

u = 1- sin3 9

du 3 . 2 -=- sm 9cos9 d9

v =cos 9

dv -=-sin 9 d9

dr -3 sin2 9 cos2 9 + sin 9 (1 - sin3 9) -= d9 cos2 9

-2 sin9 ·3 = -3 Sin 9 + -- (1- sm 9)

cos2 9

8. r=a(sin29-rcos29)

dr . - =a (2 cos 2 9-2 Sin 2 9) d9

= 2a (cos 2 9 -sin 2 9) Ans.

10. xy +cotxy=O

y + x :- (csc2 xy) ~ + x :) = 0

dy 2 dy 2 y + x dx - y esc xy - x dx esc xy = 0

dy 2 2 - (x-xcsc xy)=ycsc xy-y dx

dy -y(l - csc2 xy) y -= =-- Ans. dx x(1 -esc xy) x

= -3 sin2 9 +tan 9 sec 9 (1- sin3 9) Ans.

73

11. y=sinx

12.

1.

3.

dy --- = cosx dx

1r ..j2 a) cos 4" =-2- Ans.

. 2 X 180 114.59° = 114°35'24" b) 2 radians = 3_1416 =

cos 114°35'24" =-cos 65°24'36" =-0.4161 Ans.

tan 8 la I =I --=-- tan8

a tan 8a tan 8a

dl Ia 2 sec2 8 -=--sec 8=1 -- Ans. d8 tan8a atan8a

Self-Test 5

. -1 y =Sin 2x 2. y = tan-1 (2x-1)

dy 2 dy 2 -=

dx -J 1 - (2x)2

2 Ans.

-J1-4x2

2

4x 2 -4x + 2

y =COS-I 3x 4. y = sec-1 3x

dy -3 dy 3 - = Ans. -= dx

-J 1- 9x2 dx

3x-J9x2 -1

5. y = csc-1 (2x + 1) 6. y =coC1 (x2 -1)

dy 2 - =- --~:=:::;::== dx (2x + 1) ..J(2x + 1)2 - 1

2

(2x + 1) -J 4x2 + 4x

1 Ans.

dy -2x

dx 1 + ~2 -1)2

74

2 Ans. 2x-2x+1

Ans.

x-J9x2 -1

-2x Ans.

7. e = sin-1 G)

"9.

1 1 de 1 2 dx ;:I 1 r-;

-Y4-x2 2 4

1 Ans. .J 4 -x2

e=tan-{~~) x+1

x-1 u=--

X+ 1

du (x + 1)- (x- 1)

dx (x + 1)2

de X

2

(x + 1)2

2

1 -1 1 8. e = 3 tan 3x

Ans .

2

dx 1+c-1j

(x + 1)2 X

x 2 + 2x + 1 + x 2 - 2x + 1 (x + 1 )2

X + 1 (x + 1)2

(x + 1)2 2 2x2 + 2 X (x + 1)2 - x2 + 1 Ans.

10. y=xsin- 1 x+~

=sin -I x + __ x __ + x Ans.

~~

Self-Test 6

u = 2x -J

du _2 -2 -=-2x =­dx x2

dy -2ex dx- T Ans.

3. y = (2x2 -2x + l)e2 x

dy (4 2X 2X 2 - = x - 2)e + 2e (2x - 2x + 1) dx

2. 2 -x y =e

dy -x2 - = -2xe = -2 xy Ans. dx

= e2x (4x- 2 + 2(2x2 - 2x + I)]= e2 x (4x- 2 + 4x2 - 4x + 2) = 4x 2e2 x Ans.

75

4. Y = (3 cos x + sin x)e3x

dy . 3X 3X · 3X -=(-3smx+cosx)e +3e (3cosx+smx)=10e cosx Ans. dx

7. exsiny=2 8.

.V dy ex+ e- -= 0

dx

X

dy = -e =-e(x-y) Ans. dx eY

sinx + cosx y=

X X dy e siny + (e cosy) dx = 0 dy (cosx- sin x)ex- ex(sin x +cos x)

dy -ex siny

dx ex cosy -tany Ans.

1. y =Ln (x 2 + 2x)

dy 2x + 2 Ans.

dx x2 + 2x

2 3. y =log-

X

u = 2x -1

du -2 -2 -=-2x dx x2

dy ·= loge ( -2) dx 2 2

- X X

= xloge(-2)= -loge ;! x2 x

Ans.

dx

-2ex sinx -2 sinx =

Self-Test 7

2. y = (Lnx) 3

dy = ~ (Lnx)2 Ans. dx X

4. y =Ln cosx

dy = - sin x = - tan x Ans. dx cosx

76

Ans.

S. y=Ln(tanx+secx)

dy sec2 x + secx tanx -= dx tan x + sec x

sec x (sec x + tan x) = = secx Ans.

secx + tanx

x2 7. y=Ln --2

1 +x x2

u=~

du 2x (l + x 2)- 2x (x2 ) -= dx (l +x2) 2

dy (1 + x 2) (2x) X

dx x2 (1 + x2)2

9. e<x + y) = Ln ::

2

(1 + dy)e<x + ~ = ~ (y -x ~) dx x y2

2x

(x + y) dy (x + y) 1 1 dy e +-e =----

dx x ydx

dy fe<x. + y) + ~) = ~- e (x + y) dx \ y x

dy (ye<x + y) + 1\ 1 -xe(x + y)

dx y J x

Ans.

dy y [1 -xe<x + y)J y -xy e<x + y)

dx x [ 1 + ye<x + Y>] x +xy e<x + y)

1 d 1 -­_.!!. = 1 + - X 2x (x2 + 9) 2 dx 2

X =1+ q-;;

dy ( 1

dx= x+..Jx2+9

Ans.

dy 2 2 2x -= 2xLnx +x x­dx x2

= 2x Ln x 2 + 2x = 2x(l + Ln x 2 ) An~.

Ans.

77

1

10. y =Ln (1 +x 2 ) 2 - x tan-1 x

dy

dx

11. xy = Ln (x + y)

X X

y + x dy = _1_ (1 + ddyx) dx x +y

( 1 ) dy 1 \x- x+y dx = x+y -y

(x2 +xy- 1\dy = 1-xy-y 2

X+ Y jdx X+ y

- (t.an-1 x + _x )= x 1 + x2 1 + x2

-1 X -tan x---

1 +x2

12. y=Ln(Lnx)

dx = (Ln ~)G)= x L~ x Ans.

dy 1-xy -y2

d 2 AnL X X + xy -1

13 . y= 3 secx

1.

Lny=Ln 3secx

Ln y = sec x Ln 3

1 dy --d = (Ln 3) tanx secx )' X

dy -= y Ln 3 tanx secx Ans. dx

Or, use formula 35 directly :

dy = 3sec x Ln 3 tan x sec x Ans. dx

e 00 10° 15° 20°

r 4 2$ 2Vl 2

e r

e 240° 250° 255° 260°

r 4 2.../3 2Vl 2

14. y =xLn x

Lny = LnxLnx

Lny = (Ln x)2

1 dy 1 -- = (2 Lnx)­y dx x

dy 2y Lnx -= -- Ans. dx x

Self-Test 8

30° 90° 100° 105°

0 0 2 2.,fi

270° 330° 340° 345°

0 0 2 2.,fi

78

110°

2...[3

350° 360°

2...[3 4

r 0 2.Jf

79

3.

()

r

4.

I () 00 30° 45° 60° 90° 120° 131°48'38" 135° 150°

r 5 2 + 3y'3 2

3v'2 2+-2- 3.5 2 .o.s 0 2- 3v'2

2 2- 3y'3

2 or -0.12 or -0.60

180° 210° 225° 228°11 '22" 240° 270° 300° 315° 330° 360°

-1 2 _3v'3 2

2- 3..;2 2

or-0.60 or-0.12

() 00 15° 22°30' 30°

r 0 2 2v'2 20

(} 1120° 135° 150°

r l-2v'3 -4 -2v'3

() 240° 247°30' 255°

r 2v'3 2v'2 2

() 345° 360°

r -2 0

0

45°

4

157°30'

-20.

270°

0

0.5 2 3.5 2 + 3..;2 2

2 + 3vG 2

5

60° 67°30' 75° 90° 105° 112°30'

2../3 2v'2 2 0 -2 -2.J2

165° 180° 195° 202°30' 210° 225°

-2 0 2 2v'2 2$ 4

285° 292°30' 300° 315° 330° 337°30'

-2 -2v'2 -2J3 -4 -2v'3 -2J2

80

Self-Test 9

1. a) First find the angle that the curve makes with in radius vector, as in example 2 a). Remember

that 0 is the angle that the radius vector makes with the initial line.

r =asiu30 dr = 3a cos 30 dO

tan 1/1 = r dO = a sin 30 dO = ~ tan 30 dr 3a cos 30 dO 3

0 0 1 0 For = 60 , tan 1/1=- tan 180 = O,and 1/1 = 0°

3

1/J = 0 + 1/1 = 60° + 0° = 60° Ans.

b) r=asin30=asin180°=aXO=O Ans.

c) dr = 3a cos 30 dO

dr dO -=3acos30-dt dt

dr

dt 3a Cos 18oo X 1 3 -3a

2 = 2 a (-1) = 2 Ans.

81

2 .. a) ()

r=­tr

1 dr =- dfJ

1r

() - dfJ

r dfJ 7r 7r tan 1/1 = -- = --= fJ =- = 1.5708

dr 1 2 - dfJ

1 b) dr = - dfJ

1r

1r

dr 1 dfJ 1 = -- = - x 27r = 2 Ans.

dt 1r dt 1r

3. a) r = 1.5(1 +cos fJ) = 1.5 + 1.5 cos()

dr = -1.5 sin() dfJ

tan 1/1 = _.!_!!!!__ = 1.5 (1 + cos fJ) dfJ dr -1.5 sin fJ dfJ

-(1 +cos fJ) -(1 + cos 30°) = -{1 + 0.86603) = -3.73206

sin() sin 30° 0.5

1/1 = 180° - 75° = 105° Ans.

b) ¢ = () + 1/1 = 30° + 105° = 135° Ans.

c) dr = -1.5 sin() dfJ

dr . dfJ . o - = -1.5 SID fJ- = -1.5 SID 30 X 2.0 = (-1.5) (0.5) (2.0) = -1.5 dt dt

The radius is decreasing at 1.5 radians per second. Ans.

Self-Test 10

. y 4 J ·2 3 1. r = 5, y = 4, SID fJ = - = - , COS (} = 1 - SID (} = -

r 5 5

Use the formula on page 53 to find w

V =-rwsinfJ X

4 -120=-5 X- W=--4W

5

W=30

82

2.

2.

3 a) Vy=rwcos8=S X 30X S = 90unitsfsec Ans.

b) Angular velocity = W = 2rr n

30 = 2rr n

30 IS n = -=-rev/sec Ans.

2rr rr

a) X =10cos4t

Vx = dx = --40 sin 4t Ans. dt

b) y = 10 sin 4t

Vy dy

Ans. =- = 40 cos 4t dt

c) v = Jvx2 + v 2 y Jc--40 sin 4t)2 + (40 cos 4t)2 )I600(sin2 4t + cos2 4t)

.JI600 = 40 Ans.

d) ax d V x d(--40 sin 4t)

= -- = =-160 cos 4t Ans. dt dt

e) ay d Vy d(40 cos 4t)

= -- = = -160 sin 4t Ans. dt dt

d8 d(4t) 0

g) w = - = -- = 4 radums per second Ans. dt dt

Self-Test 11

a) d8

w = - = 12t2 - 24t = 12 X 9 - 24 X 3 = 36 radians per sec Ans. dt

b) V = rX w = 2 X 36 = 72unitspersec Ans.

c) dw

a = - = 24t - 24 = 24 X 3 - 24 = 48 radians per sec2 Ans. dt

d) at = r X a = 2 X 48 = 96 units per sec2 Ans.

v 2t3 - 6 2 X 8-6 = 5 radians per sec Ans. a) w =- =

r r 2

b) dw 6t2 6X4

12 radians per sec2 Ans. a =-=- -2- = dt 2

83

c) at = r X a = 2 X 12 = 24 units per sec2 Ans.

d) ac = -rw2 =-2 X 25 = -50

aR = )at2 + ac2 = J242 + (-50)2 = .j3Qi6 = 55.46,or 55.5 units per sec2 Ans.

1. y = sinh 2x

dy - = 2 cosh 2x Ans. dx

3. y = tanh 2x 3

5. y = csch (~)

7.

- =-- csch - coth -dy 1 (X) (X) dx 2 2 2

1 . 1 y=-smh2x--x

4 2

dy 1 1 -=- cosh2x-­dx 2 2

1 = 2 (cosh 2x-1) Ans.

1. y = sinh - 1 (3x)

dy 3 Ans. -=

dx

Ans.

Self-Test 12

2. y = cosh2 (tan x) = u2

u = cosh (tan x)

du 2 . - = sec x smh (tan x) dx

dy - = 2u = 2 cosh (tan x) du

dy22"() - = sec x smh tan x cosh (tan x) Ans. dx

4. y = cosh2 Sx - sinh2 Sx

dy = 0 An s. dx

6. y = Ln (cosh 3x)

dy = __ l_ X 3 sinh 3x = 3 tanh 3x Ans. dx cosh 3x

8. coth 2y = tan 3x

(-2 csch2 2y>(ix)= 3 sec2 3x

dy - = ----=-dx

3 sec2 3x Ans.

-2 csch2 2y

Self-Test 13

2. y = tanh - 1 (sin x) = tanh - 1 u

84

. du u = sm x, - = cos x

dx

dy X cosx -=

dx

cosx

cos2 x cosx secx Ans. =---

3. y = coth - 1 {~) = coth - 1 u

7.

9.

I -1 u=-=x

X

du _2 -1 - = -x =-dx x2

dy -1 -=----x--dx x2 - x2

x2

-1 = --- =--- Ans.

dy -2x -2 -=--.:::....--=----dx .r----:­

x2v1 + x4

y = cosh-1 &'

dy ex -= dx

Je2x - 1 Ans.

sinh -l (3x) = sinh (3y)

3 dy 3 cosh (3y)-

dx J1 + 9x2

dy dx

J1 + 9x2 cosh (3y)

Ans.

4. y = sech-1 (cosx) = sech-1 u

Ans.

8.

85

du . u = cosx, dx = -smx

dy -sin x dx =- --=;:=:;;:::-

cos x .J1 - cos2 x

sinx = =--

COS X sin X COS X

X u=-2'

du 1 dx = 2'

dy -=----X dx

y = tanh-1 (4x2 )

2

Ans.

dy 8x Ans.

dx - 16x4

= secx Ans.

10.. y = 2 tanh-1 ~an~)= 2 tanh-1 u

X u= tan 2,

dy 2 -=

du dx

dx 2 X - tan -

2

X

1 2 X

2 sec 2

2 X sec

2 2

~"'X . 2

Sin 1 2

--x-+ 2 X cos2 2 cos

2

secx Ans. cosx

sec2 X

2

1 - tan 2 X

2

X) 2 - 2 X

cos 2-

86

( . ,x ) Sin -1 2

--+ 1---2 X X

cos 2 cos2 2

. 2 X Sin

2 cos (2 X D

EXAMINATION 6618C

Calculus: Function and Use, Part 3 When you feel confident that you have mastered the material in this study unit,

complete the following examination. Then submit only your answers to school

headquarters for grading, using one of the examination answer options de­

scribed in your first shipment. Send your answers for this examination as soon

as you complete it. Do not wait until another examination is ready.

Questions 1-15: Select the one best answer to each question.

dy 1. Calculate dx when xy3 + y = 3x.

A.-3-y3 + I

B.-3-y3- I

dy (3x + 1)3 2. Calculate- when y2 = -----::--~

dx 9x + 2 ·

-3(3x + 1)2 A. -::--:::------:~

2y(9x + 2)2

-3(3x + 1)2 B.

y 2(9x + 2)2

c. 3 + y3

I + 3xy2

D. 3- y3

l + 3xy2

9(3x + 1)2 (6x + I) c. --~~-~~-

2y(9x + 2)2

9(3x + 1)2 (6x + I) D. ---::-=----::~-

y2(9x + 2)2

3. Given the functions y = 3x3 + 6x and x = 2t2 + t, find ~.

A. 9t2 + 6

B. (9x2 + 6)(4t + I)

4. Find dx when y = cosh (bx2).

A. 2bx cosh (bx2)

B. -2bx cosh (bx2)

S. Find : when y = Ln (sinh 2x).

A. 2 cosh 2x

B. 2 coth 2x

9x2 + 6 c. 4t + I 4t + I

D.---9x2 + 6

C. 2bx sinh (bx2)

D. -2bx sinh (bx2)

C. 2 sech 2x

D. 2 csch 2x

6. Find : when sinh 3y = cos 2x.

A. -2 sin 2x

B. -2 ~in 2x smh 3y

2 2x C. --tan(-)

3 3y

D. _ 2 sin 2x 3 cosh 3y

7. Find the derivative of y = cos (x2) with respect to x.

A. -sin (2x) C. - sin (2x) cos (x2)

B. -2x sin (x2) D. -2x sin (x2) cos (x2)

8. Find the derivative of y = sin2 (4x) cos (3x) with respect to x.

A. 8 sin (4x) cos (3x)- 3 sin2 (4x) sin (3x)

B. 8 cos (4x) cos (3x) - 3 sin2 (4x) sin (3x)

C. 8 sin (4x) cos (4x) cos (3x) - 3 sin2 (4x) sin (3x)

D. 8 sin (4x) cos (4x) cos (3x) - 3 sin2 (4x) sin (3x) cos (3x)

9. Find the derivative of y = ebx2 with respect to x.

A. bebx2 C. 2 bxebx2

B. bxebx2 D. bx2 ebx2

10. Calculate dy when y = (x2 + 2)e 4x as the derivative of a product, letting dx

u = x2 + 2 and v = e4x. Which of the following is a step in your solution?

du dv A.-= 2x + 2 C.-= 4xe3x

dx dx

B. du = 2x dx

dv D.-= 4xe 4x

dx

11. Which of the following represents : when y = e-2x sec (3x)?

A. 3e -2x sec (3x) tan (3x) - 2e -2x sec (3x) B. 3e -2x sec (3x) tan (3x) - 2xe -2x sec (3x)

C. 3e-2x sec (3x) tan (x) -2 e-2x sec (3x)

D. 3e-2x sec (3x) tan (x) - 2xe-2x sec (3x)

12. Calculate : if y = In (2x3 + 3x).

1 A. 2x3 + 3x

1 B. 6x2 + 3

2

2x3 + 3x C. 6x2 + 3

6x2 + 3 D.---

2x3 + 3x

dy ..!. 13. Calculate dx if Ln (x + y) = eY.

X X

eY xy+eY y2+y2 A.~~~~~--~

X X

e'Y xy+eYx2+y2

X X

eY xy+eY y2-y2 B. ..:.....,x:--=--.::.......,.x_;__..::~

eY xy-eY x2+y2

X X

eY xy+eY y2+y2 c. --..,--=----=~--=-~ !...

eY xy-eY x2+y2

X X

D. e Y xy + e Y y2 - y2 X !_

eY xy+eY x2+y2

14. Given the curve that is described by the equation r = 3 cos 9, find the angle that the tangent line makes with the radius vector when 9 = 120°.

A. 30° B. 45°

c. 60° D. 90°

15. A line rotates in a horizontal plane according to the equation 9 = 2t3- 6t, where (J is the angular position of the rotating line, in radians, and t is the time, in sec­onds. Determine the angular acceleration when t = 2 sec.

A. 6 radians per sec2 C. 18 radians per sec2 B. 12 radians per sec2 D. 24 radians per sec2

3