chapter 2 differentiation-2016

8/18/2019 Chapter 2 Differentiation-2016

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Example: Find the slope of the curve f ( x) = x2

Solution: We start with a point Q(1, 1) which is near P (2, 4):

at the point (2,4), using a numerical method.

The slope of PQ is given by

Now we move Q further around the curve so it is closer to P

The slope of PQ is now given by:

. Let's use Q(1.5, 2.25)which is closer to P (2, 4):

We see that this is already a pretty good approximation to the tangent at P

Now we move Q even closer to P, say Q(1.9, 3.61).

, but notgood enough.

Now we have:

o

We can see that we are very close to the required slope. Now if Q is move to (1.99,

3.9601), then slope PQ is 3.99. If Q is (1.999, 3.996001), then the slope is 3.999.

Figure:2.3 (b)

Figure:2.3 (c)

Figure:2.3 (a)


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Clearly, as x → 2, the slope of PQ → 4. But notice that we cannot actually let x= 2,since the fraction for m would have 0 on the bottom, and so it would be undefined. Wehave found that the rate of change of y with respect to x is 4 units at the point x

We will now extend this numerical approach so that we can find the slope of any

continuous curve if we know the function. We will see an algebraic approach that can be used for most functions.

= 2.

2.2 THE SLOPE

Most functions we are interested in are not straight lines. We cannot define a gradientof a curved function in the same way as we can for a line. In order for us to understandhow to find the gradient of a function at a point, we will first have to cover the idea oftangency . A tangent is a line which just touches a curve at a point, such that the angle

between them at that point is zero.

ON A FUNCTION

Consider the following four curves and lines:

(i) (ii)

i. The line L crosses, but is not tangent to C at P .ii. The line L crosses, and is tangent to C at P .

iii. The line L crosses C at more than one point, but is tangent to C at P .iv. There are many lines that cross C at P , but none are tangent. In fact,

this curve has an undefined tangent at 'P .

A secant is a line drawn though two points on curve. We can construct a definition ofthe tangent as the limit of a secant of the curve drawn as the separation between the

points tends to zero. Consider the diagram below.

Figure:2.4 (a) Figure:2.4 (b)

Figure:2.4 (c) Figure:2.4 (d)

http://en.wikibooks.org/wiki/Image:Tangency_Example_2.svghttp://en.wikibooks.org/wiki/Image:Tangency_Example_1.svghttp://en.wikibooks.org/wiki/Image:Tangency_Example_4.svghttp://en.wikibooks.org/wiki/Image:Tangency_Example_3.svg


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As the distance h tends to zero, the secantline becomes the tangent at the point x 0 .The two points we draw our line through

are:

( , ( ))o oP x f x and( , ( ))o oQ x h f x h+ +

As a secant line is simply a line and weknow two points on it, we can find itsslope, m:

This expression is called the difference quotient . Note that h can be positive ornegative.

Now, to find the slope of the tangent line at o x , we let h zero. We must find thelimit of the above expression as h tends to zero:

Let

2.3 DERIVATIVE OF A FUNCTION AT A SPECIFIC POINT.

The differentiation or derivative is the formula m tan

Or or

(slope of tangent line) on a curveat a specific point. So

Figure:2.5

0 0( , ( ))

( ) ( )lim

0o o

x f x

f x h f xdy

hdx h

+ −=

→

http://en.wikibooks.org/wiki/Image:Tangent_as_Secant_Limit.svghttp://en.wikibooks.org/wiki/Calculus/Limitshttp://en.wikibooks.org/wiki/Calculus/Limitshttp://en.wikibooks.org/wiki/Image:Tangent_as_Secant_Limit.svghttp://en.wikibooks.org/wiki/Calculus/Limits


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This is the definition of the derivative of a function at o x x= . If the limit exists we

say that f is differentiable at o x x= and its derivative at o x x= iso x x

dydx =

The definition of the derivative of a function at any x if the limit exist, is :

( ) ( )lim

0

d y f x h f x

hdx h

+ −=

→ is called

first principles of differentiation.

Some common notation for derivativeare:

Eample : Finddy

dx from first principles if y = 2 x2+ 3 x

Solution:

.

Let f ( x) = 2 x2+ 3 x

so

We now need to find dydx

:

We have found an expression that can give us the slope of the tangent anywhere onthe curve.If xIf

= -2, the slope is 4(-2) + 3 = -5 x

If

= 1, the slope is 4(1) + 3 = 7

x = 4, the slope is 4(4) + 3 = 19

Figure:2.6


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We can see that our answers are correct when we graph the curve (which is a parabola ) and observe the slopes of the tangents.

This is what makes calculus so powerful. We can find the slope anywhere on thecurve (i.e. the rate of change of the function anywhere).

Example :

a. Find y' from first principles if y = x 2 + 4 x

b. Find the slope of the tangent where

.

x = 1 and also where

c. Sketch the curve and both tangents.

x = -6.

Solution:

a. Note: y' means "the first derivative". This can also be written dy/dx .

Now f ( x) = x2

+ 4 x

So

b. When x = 1, m =When x = -6, m =

c. Sketch:2.3.1 Definition A function f is differentiable at a if ( ) f a′ exists. It isdifferentiable on an open interval (a,b) [or (a, ∞ ) or ( ∞ ,a) or ( ∞ , ∞ )] if it f is differentiable at every number in the interval

Example: Where is the function ( ) f x x= differentiable?

Figure:2.7

Figure:2.8

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2.3.2 Theorem : Differentiable Implies Continuous

If f is differentiable at a , then f is continuous at a .

(The converse of this theorem is false)

In the next section, we will see some (much simpler) rules for differentiation. We

won't use "differentiation from first principles" very often, but it is good to have anunderstanding of where differentiation comes from and what it can do for us.

Figure:2.9


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2.4 HOW CAN A FUNCTION FAIL TO BE DIFFERENTIABLE

We saw example 3.31 ( ) f x x= is not differentiable at x= 0 , this is becausethe graph changes direction abruptly (sharply) when x=0 . In general, f is not

differentiable at x= a, if:(i) the graph of a function f has a sharp corner (cusp).

(ii) f not continuous at a .

(iii) the graph of a function f has a vertical tangent line at x=a .

Example: Consider ( ) y g x= . Find x, where ( ) y g x= not differentiable.

Example: Find x at which f is not differentiable.

(i) ( ) 2 f x x= − (ii)2

( ) 2 f x x x=

− (iii) 3( ) f x x=

Figure:2.10

Figure:2.11


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2. 5 RATES OF CHANGE

Suppose y is a quantity that depends on another quantity x. Thus y is a function of xand we write y= f (x). If x change from x1 to x2

So, the difference quotient

, then the change in x is

is called the average rate of change of y with respect to x.

2 10 0 2 1

( ) ( )lim lim x x

f x f x y x x x∆ → ∆ →

−∆ =∆ −

is called rate of change or instantaneous rate of change

We can find the

.

rate of change

Temperature change at a particular time

of one variable compared to another in this examples:

Velocity of a falling object at a particular timeCurrent through a circuit at a particular timeVariation in stock market prices at a particular timePopulation growth at a particular timeTemperature increase as density increases in a gas

Later, we will see how to find these rates of change by differentiating a function andsubstituting a value.

Example:

1. The area of a circle is related to its diameter d by equation2

4 A d

π =

How fast does the area change with respect to the diameter when the diameter is 10m.2. Population growth rate…

Solution: 1. The rates of change of the area with respect to the diameter is

2 1

2 1

( ) ( ) f x f x y x x x

−∆=

∆ −


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2.5.1 Motion along a Line

Suppose that an object is moving along a coordinate line, usually horizontal or vertical,

so that we know its position s on that line as a function of time t, ( )s f t = .

Displacement ( ) ( )s f t t f t ∆ = + ∆ − Velocity =Acceleration =Speed =

Example: The free fall of a heavy ball bearing released from rest at time t =0 sec

with equation24.9s t = .

(a) How many meters does the ball fall in the first 3 sec.(b) What is its velocity, speed and acceleration when t = 3?

2.6 SOME RULES FOR DIFFERENTIATION

1) Constant Rule: If ( ) f x c= , then

2) Power Rule: ( ) 1n nd x nxdx−= , n positive

3) Constant multiple: If ( ) y cf x= , then

4) Sum and Difference: ( )( ) ( ) ( ) ( )d

f x g x f x g xdx′ ′± = ±

Example:

Figure:2.12


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2.6.1 Derivatives of product and quotient of two Functions

If ( ) and ( )u x v x are differentiable at every x, then

1) ( ) d dv du

uv u vdx dx dx

= +

2) ( ) 2 u

du dvvd dx dxuv

dx v

+=

2.6.2 The Chain RuleUseful for derivative of composite functions.

If ( ) f u is differentiable at the point ( )u g x=

and ( )g x is differentiable at x , then the composite function ( )( ) ( ( )) f g x f g x= is differentiable at x,and

( ) ( ) ( ( )) ( ) f g x f g x g x′ ′ ′=

In Leibniz’s notation, if ( ) y f u= and ( )u g x= , then

dy dy dudx du dx

= ,

Where dy/du is evaluated at ( )u g x= .

Example:

2.6.3 Higher Order of Differentiation

Assumed f can be differentiated as often as necessary, the second derivative of f

is [ ]2

(2)2( ) ( ) ( )

d f d f x f x f x

dxdx′′ ′= = =

For integer 1n ≥ , nth derivative is

( ) ( 1)( ) ( )n

n nn

d f d f x f x

dxdx

− = =

Example: Find the third derivative of 3( ) 4 5 9 f x x x= − +


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2.7 LOGARITHMIC DIFFERENTIATION

The calculation of derivatives of complicated functions involving products,quotients or powers can be simplified by taking logarithms.

Steps:1. Take logarithms of both sides of equation and use properties of logarithm

to simplify.2. Differentiate with respect to x.

3. Solve for dydx

Example: Find

2.8 IMPLICIT DIFFERENTIATION

These following equations define an implicit relation between the variables x and y.

3 3 2 39 0 and sin 0 x y xy y xy x+ − = + − =

We cannot write an equation ( , ) 0F x y = in the form ( ) y f x=


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2. 9 DERIVATIVES OF FUNCTIONS

2.9.1 Derivatives of Trigonometric Functions

y dydx

sin x cos xcos x ̶ sin xtan x sec 2 xcosec x ̶ cosec x cot xsec x sec x tan xcot x ̶ cosec 2 x

Example:

Table :2.1


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2.9.2 Derivatives of Exponential Functions:

y dy

dx

e e x x ( )f x e ( ) ( )′

f x e f x

x a lnx a a , a - constant( )f x a ( ) ( )′ ln

f x a f x a

Example:

Table :2.2


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2.9.3 Derivatives of Logarithmic Functions:

y dy

dx

ln x 1x

( )ln f x ( )( )

f x

f x

′

loga x 1lnx a

( )loga f x ( ) ( )1 1

lnf x

a f x ′⋅ ⋅

Example:

Table :2.3


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2.9.4 Derivatives of Inverse Trigonometric Functions

y dy

dx

1sin x − 21

1 x − , -1 < x


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2.10 RELATED RATE

The problems of finding a rate of change from other known rate of change is called arelated rate problem.

EXAMPLE

1. A tank of water in the shape of a cone is leaking water at a constant rate of

. The base radius of the tank is 5 ft and the height of the tank is 14 ft.

(a) At what rate is the depth of the water in the tank changing when the depth of thewater is 6 ft?(b) At what rate is the radius of the top of the water in the tank changing when thedepth of the water is 6 ft?Solution

As we can see, the water in the tank actually forms a smaller cone with the samecentral angle as the tank itself. The radius of the “water” cone at any time is given

by r and the height of the “water” cone at any time is given by h . The volume ofwater in the tank at any time t is given by,

and we’ve been given that 2dV dt

= − .

(a) At what rate is the depth of the water in the tank changing when the depth of thewater is 6 ft?

For this part we need to determinedhdt

when h=6 .

From the figure,5

14r h

= . Thus5

14r h=

Then

231 5 25

3 14 588V h h hπ π

= =

The only formula that we’ve got that will relate the volume to the height also includethe radius and so if we were to differentiate this with respect to t we would get,

Fi ure:2.13


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dV dV dhdt dh dt

=

So, in this equation we know = -2and h = 6 , and want to find h ′ .

23(25)2 588

dhh dt

π

− =

When h = 6, 275

(6) 0.13862(588)

dhdt

π = − ≈ −

The height is decreasing at a rate of 0.1386 ft/hr.

( b) At what rate is the radius of the top of the water in the tank changing when thedepth of the water is 6 ft?

In this case we are asking for dr dt

Recall from the first part that we have,

2.11 INDETERMINATE FORMS AND L’HOSPITAL RULE

Example:

Find the limit by using L’Hospital rule if2

212lim

x x x

x x→+ −

−


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2.11.1 Indeterminate Forms ∞/∞, ∞·0, ∞−∞

Example:

(i) Find the limit of this form.

(ii) Find the limit of this . 0 form.

Strategy: If ( ) f x and ( )g x are differentiable functions then rewrite the

product fg as a quotient;( ) ( )

( ). ( )1 / ( ) 1/ ( )

f x g x f x g x

g x f x= = and apply

L’Hospital rule.

(a)

(b)

(ii) Find the limit of this form.

Strategy: If ( ) f x and ( )g x are differentiable functions then convert thedifference f-g into a quotient ( using a common denominator, rationalization orcommon factor) and apply L’Hospital rule.

/∞ ∞


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2.11.2 Indeterminate Forms 1∞

, 0 0 , ∞ 0

Sometime handling by taking logarithm of the functions.

Strategy: If ( ) f x and ( )g x are differentiable functions in the form ( )( ) g x f x then

(a) Take natural log. Let( )

( )g x

y f x= ln ( ) ln ( ) y g x f x=

(b) Followed by strategy 1 or 2.

Example: Find the limit

chapter 2 differentiation-2016

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