cal1 iu slides(2ndsem09-10) chapter 2 sv

7/27/2019 Cal1 IU SLIDES(2ndSem09-10) Chapter 2 SV

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Chapter 2 DIFFERENTIATION

Contents

[1.] Tangent Lines and Their Slopes

[2.] Derivatives

[3.] Differentiation Rules[4.] Rates of Change in Natural and Social Sciences

[5.] Derivatives of Trigonometric Functions

[6.] Chain Rule and Implicit Differentiation

[7.] Derivatives of Inverse Functions

[8.] Linear Approximation and Applications

Dr. Nguyen Ngoc Hai CALCULUS I
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2.1 TANGENT LINES AND THEIR SLOPES

Differential calculus is concerned with how one quantity changes inrelation to another quantity.

Its central concept of differential calculus is the derivative.

Questions: How can we define the tangent line to a

graph and how can we compute its slope?

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Let C be the graph of y = f(x) and let P be the point (a, f(a)) on

C. Consider a nearby point Q(x, f(x)), where x = a, and the linethrough P and Q, which is called a secant line to the curve. Thisline rotates around P as Q moves along the curve.

Slope of secant line through P and Q = fx

= f(x) f(a)x a

where f = f(x) f(a) and x = x a.

The expressionf(x) f(a)

x ais called the difference quotient.

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Let Q approach P along the curve C by letting x approach a.Then the tangent line is the limiting position of the secant line PQas Q approaches P.

Definition 1.1 The tangent line to the curve y = f(x)at the point P = (a, f(a)) is the line through P with slope

m = limxa

f(x) f(a)x a

provided that this limit exists.

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Let h = x a. Then x = a + h and the slope of the secant line PQis

mPQ =f(a + h)

f(a)

h .

As x a, h 0 and so the slope of the tangent line is

m = limh0

f(a + h) f(a)h

Example 1.1 Consider the graph of f(x) = 3

x = x1/3. Let ustry to calculate the limit of the difference quotient for f at x = 0:

limh0

f(0 + h) f(0)h

= limh0

h1/3

h= lim

h01

h2/3=

Hence although the limit does not exist, the slope of the secantline joining the origin to another point Q on the curve approaches

infinity as Q approaches the origin from either side.Dr. Nguyen Ngoc Hai CALCULUS I
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Definition 1.2 If f(x) is continuous at the pointP = (a, f(a)) and if either

limh0

f(a + h) f(a)h

= or limh0

f(a + h) f(a)h

= ,

then the vertical line x = a is tangent to the graphy = f(x) at P.

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Definition 1.3 The slope of a curve C at a point P isthe slope of the tangent line to C at P if such a tangent lineexists. In particular, the slope of the graph of y = f(x) at the

point (a, f(a)) is

limh0

f(a + h) f(a)h

.

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Definition 1.4 If a curve C has a tangent line L at pointP, then the straight line N through P perpendicular to L iscalled the normal to C at P.

If L is horizontal, then N is vertical; if L is vertical, then N is

horizontal. If L is neither horizontal nor vertical, then the slope ofN is the negative reciprocal of the slope of L:

slope of the normal =1

slope of the tangent

Example 1.2 Find equations of the straight lines that aretangent and normal to the curve y =

x at the point (4, 2).


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2.2 DERIVATIVES2.2.1 DERIVATIVES. RATES OF CHANGE

Definition 2.1 The derivative of a function f at a

number a, denoted by f(a), is the limit of the differencequotients

f(a) = limh0

f(a + h) f(a)h

if this limit exists. When the limit exists, we say that f isdifferentiable at a. The process of computing the derivativeis called differentiation.

If we writex

=a

+h

, thenh

=x

a

andh

approaches 0 if andonly if x approaches a. Therefore, an equivalent definition of thederivative is

f(a) = limxa

f(x) f(a)x a


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Finding f(a) from the definition of derivative

The algebraic steps to calculate f(a):

Step 1. Find f(a) and f(a + h).

Step 2. Find and simplify f(a + h) f(a) .Step 3. Divide by h to get f(a+h)f(a)

h.

Step 4. Take the limit as h 0.


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Geometrical Meaning of the Derivative

The tangent line to the curve y = f(x) at the point P(a, f(a)) hasslope

m = limh0

f(a + h) f(a)h

= f(a)

So

The tangent line to y = f(x) at (a, f(a)) is the line through(a, f(a)) whose slope is equal to f(a), the derivative of f at a.

Therefore, an equation of the tangent line to the curve y = f(x)at the point (a, f(a)) is

y = f(a) + f(a)(x a)


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Suppose y is a quantity that depends on another quantity x. Thus,y is a function of x: y = f(x).

If x changes from x1 to x2, then the change in x (also called theincrement of x) is x = x2 x1.

The corresponding change in y is y = f(x2) f(x1).

The difference quotient

y

x =

f(x2)

f(x1)

x2 x1is the average rate of change of y with respect to x over theinterval [x1, x2].


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The limit of average rates of change as x 0 is called the(instantaneous) rate of change of y with respect to x at x = x1:

limx0y

x = limx2x1f(x2)

f(x1)

x2 x1

Thus,

The derivative f(a) is the instantaneous rate of change of

y = f(x) with respect to x when x = a.


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2.2 DERIVATIVES2.2.2 DERIVATIVE AS A FUNCTION

Definition 2.2 The derivative of a function f is thefunction f whose value at x is

f(x) = limh

0

f(x + h) f(x)h

if this limit exists.

The domain of f is the set

{x| f(x) exists}and may be smaller than the domain of f.


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We extend the definition to allow for a right derivative at x = a

and a left derivative at x = b:

f+(a) = limh0+

f(a + h) f(a)h

, f(b) = limh0

f(b+ h) f(b)h

.

We say that f is differentiable on [a, b] if f(x) exists for all x in(a, b) and f+(a) and f

(b) both exist.

In general,

f is differentiable on an interval I if it is differentiable atevery number in the interval.

If f has a derivative at every point of its domain, we call fdifferentiable.


2 2 DERIVATIVES
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Right and left derivatives may be defined at any point of a

functions domain.

Remark:

f is differentiable at x if and only if the one-sided derivatives

f+

(x) and f

(x) exist and are equal.

f(x) f+(x), f(x) and f+(x) = f(x)

Example 2.1 Find the derivative of f(x) = |x|.

Example 2.2 Find the derivative of f(x) =

x. State thedomain of f.


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Solution

f(x) = limh0 x + h xh

= limh0

(

x + h x)(x + h + x)h(

x + h +

x)

= limh0(x + h)

x

h(x + h + x) = limh0h

h(x + h + x)= lim

h01

x + h +

x=

1x +

x

=1

2

x.

At x = 0,

limh0+

0 + h 0

h= lim

h0+

h

h= lim

h0+1

h= .

Thus f is not differentiable at x = 0 and the domain of f is

(0, ). This is smaller than the domain of f, which is [0, ).Dr. Nguyen Ngoc Hai CALCULUS I

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Other Notations

The most common notations for the derivative of a functiony = f(x), besides f(x), are

y =dy

dx =df

dx = Dx(f) =d

dx(f)

If we want to indicate the value of a derivative at a specificnumber a, we use the notations

Dxyx=a

= dydx

x=a

= dfdx

x=a

= ddx

f(x)x=a

which are synonyms for f(a).


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2.2 DERIVATIVES2.2.3 CONTINUITY AND DIFFERENTIABILITY

Theorem 2.1 If f is differentiable at a, then f iscontinuous at a.

The converse of this theorem is false. That is, there are

functions that are continuous but not differentiable.

It follows from Theorem 2.1 thatA function can never have a derivative at a point ofdiscontinuity.


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2.2 DERIVATIVES2.2.3 CONTINUITY AND DIFFERENTIABILITY

When does a Function Fail to be Differentiable?

At any discontinuity a function fails to be differentiable.

If the graph of a function f has a corner in it, then the graphof f has no tangent at this point and f is not differentiablethere.

If f is continuous at a but the curve has a vertical tangent line

x = a then f is not differentiable at this point becauselimxa |f(x)| = .


2 3 DIFFERENTIATION RULES
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2.3 DIFFERENTIATION RULES

Constant Rule

The derivative of a constant function is 0:

d

dx(c) = 0

Constant Multiple Rule

If u is a differentiable function of x, and c is a constant, then

d

dx(cu) = c

du

dx

The derivative of a constant times a function is the constant

times the derivative of the function.


2 3 DIFFERENTIATION RULES
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Sum and Difference Rule

If u and v are differentiable functions at x, then

(u+ v)(x) = u(x) + v(x)

(u

v)(x) = u(x)

v(x)

The derivative of a sum is the sum of the derivatives. The

derivative of a difference is the difference of the derivatives.

Example 3.1 Does the curve y = x4 2x2 + 2 have anyhorizontal tangents? If so, where?


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Product RuleIf u and v are differentiable functions at x, then

(uv)(x) = u(x)v(x) + v(x)u(x)

The derivative of a product of two functions is the first

function times the derivative of the second, plus the second

function times the derivative of the first.

Example 3.2 Find the derivative of (x2 + 1)(x3 5).


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3 O U S

Power Rule

If n is a positive integer, then

d

dx(xn) = nxn1

d

dx

xexponent

= exponent xexponent1

Example 3.3

d

dxx3 = 3x2,

d

drr2 = 2r,

d

dtt48 = 48t47.


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Quotient Rule

If u and v are differentiable functions at x, and if v(x)

= 0,

then uv

(x) =

v(x)u(x) u(x)v(x)[v(x)]2

The derivative of a quotient is the denominator times the

derivative of the numerator, minus the numerator times thederivative of the denominator, all divided by the square of the

denominator.

In particular,1

v

(x) = v

(x)[v(x)]2

Example 3.4 Find the derivative of y = x21

x2+1.


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Power Rule

If n is a negative integer and x = 0, thend

dx(xn) = nxn1.

More generally,

For any real number ,

d

dx(x) = x1


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Second and Higher Order Derivatives

If f is a differentiable function, then its derivative f may alsohave a derivative of its own. If so, (f) is called the secondderivative of f and denoted by f. We refer to f as the zerothderivative and f as the first derivative. We also write thesecond derivative of y = f(x) as

y =dy

dx=

d

dx

dydx

=

d2y

dx2

If f is differentiable, its derivativef = y =

dy

dx=

d3y

dx3

is the third derivative of f.


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In general, the nth derivative of f is the derivative of the

(n 1)st derivative and denoted by

y(n) = f(n)(x) =dny

dxn=

d

dxy(n1)

Example 3.5 Find the nth derivative of the functiony = x3 4x2 + 10.

Example 3.6 Calculate the first four derivatives of y = x1.Then find the pattern and determine a general formula for y(n).


2.4 RATE F HAN E IN THE NATURALAND SOCIAL SCIENCES
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AND SOCIAL SCIENCES2.4.1 VELOCITY

Suppose we have a body moving along a coordinate line and weknow that its position at time t is s = f(t). In the interval fromany time t to the slightly later time t + t, the body moves fromposition s = f(t) to position s + s = f(t + t). The bodys netchange in position, or displacement, for this short time interval is

s = f(t + t) f(t).

Definition 4.1

If a body moves along a line from position s = f(t) to

position s + s = f(t + t), the the bodys averagevelocity for the time interval from t to t + t is

vav =displacement

travel time=

s

t=

f(t + t) f(t)t


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Definition 4.2

Instantaneous velocity (velocity) is the derivative ofposition with respect to time. If the position function of thebody moving along a line is s = f(t), the bodysinstantaneous velocity at time t is

v =ds

dt

= lim

t0

f(t + t) f(t)

t


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Motion under the Influence of Gravity

Galileo discovered that the height s(t) and velocity v(t) of anobject tossed vertically in the air are given as function of time bythe formulas

s(t) = s0 + v0t1

2 gt2

, v(t) =ds

dt = v0 gtThe constants s0 and v0 are the initial values:

s0 = s(0) is the position at time t = 0.

v0 = v(0) is the velocity at time t = 0.

g is the acceleration due to gravity on the surface of theearth, with value

g = 9.80 m/s2 = 32 ft/s2

The maximum height is attained when v(t) = 0.


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Example 4.1 A dynamite blast blows a heavy rock straight upwith a launch velocity of 160 ft/sec. It reaches a height ofs = 160t 16t2 ft after t sec.(a) How high does the rock go?(b) How fast is the rock traveling when it is 256 ft above theground on the way up? on the way down?


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Speed

Definition 4.3 Speed is the magnitude of the velocity.

speed = |velocity|

Example 4.2 Suppose a person standing at the top of abuilding 112 feet high throws a ball vertically upward with aninitial velocity of 90 ft/sec.

(a) Find the balls height and velocity at time t.

(b) When does the ball hit the ground and what is its impactspeed?

(c) When is the velocity 0? What is the significance of this time?

(b) How far does the ball travel during its flight?


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Acceleration

In studies of motion, we usually assume that the bodys positionfunction s = f(t) has a second derivative as well as a first. Thefirst derivative gives the bodys velocity as a function of time; the

second derivative gives the bodys acceleration.

Definition 4.4 Acceleration is the derivative of velocity:

a =

dv

dt =

d2s

dt2 .

So

acceleration is the second derivative of the position.


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If a(t) > 0, the velocity is increasing. This does not necessarymean that the speed is increasing. The object is speeding up onlywhen the velocity and acceleration have the same sign.

If velocity is and acceleration is then object is and speed is

positive positive moving forward increasingpositive negative moving forward decreasingnegative positive moving backward decreasing

negative negative moving backward increasing


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AND SOCIAL SCIENCES2.4.2 OTHER RATES OF CHANGE

Definition 4.5 The average rate of change of a functionf(x) over the interval from x to x + h is

Average rate of change =f(x + h) f(x)

h

The (instantaneous) rate of change of f at x is thederivative:

Rate of change of f at x = f(x) = limh0 f(x + h) f(x)h

provided that the limit exists.


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Example 4.3 Let A = r2 be the area of a circle of radius r.

(a) Calculate the rate of change of area with respect to radius.(b) Compute dA/dr for r = 2 and r = 5, and explain why dA/dr islarger at r = 5.


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The Effect of a One-Unit Change

For small values of the increment h, the difference quotient is closeto the derivative itself:

f(a) = limh0f(a + h)

f(a)

h f(a + h)

f(a)

h .

In some applications, the approximation is already useful withh = 1:

f(a)

f(a + 1)

f(a)

In other words,

f(a) is approximately equal to the change in f caused byone-unit change in x when x = a .


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Example 4.4 Researchers have determined that the body massof yearling bighorn sheep on Ram Mountain in Alberta, Canada,can be estimated by

M(t) = 27.5 + 0.3t

0.001t2,

where M(t) is the mass of the sheep (in kg) and t is the numberof days since May 25.

(a) Find the average rate of change of the weight of a bighorn

yearling between 70 and 75 days after May 25.(b) Find the (instantaneous) rate of change of weight for a

yearling sheep whose age is 70 days past May 25.


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We now consider some other rates of change in Physics.

A current exists whenever electric charges move. If Q is the netcharge that passes through a surface during a time period t,then the average current during this time interval is defined as:

Average Current = Qt

= Q Q0t t0

If we take the limit of this average current over smaller and smallertime intervals, we get what is called the current I at a given time

t0:I = lim

tt0Q Q0

t t0 =dQ

dt

Thus, the current is the rate at which charge flows through asurface.


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If a given substance is kept at a constant temperature, then its

volume V depends on its pressure P. The rate of change of volumewith respect to pressure is the derivative dV/dP. As P increases,V decreases, so dV/dP < 0. The compressibility is defined by

Isothermal Compressibility = = 1

V

dV

dP

Velocity, compressibility, and current are not the only rates ofchange important in physics. Others include:

Density Power (the rate at which work is done) Rate of heat flow Temperature gradient (the rate of change of temperature with

respect to position)

Rate of decay of a radioactive substance in nuclear physics


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Derivatives in Economics

Marginal Cost Suppose c(x) is the total cost of producing xunits of a particular commodity. It costs more to produce x + hunits, and the cost difference, divided by h, is the average increasein cost per unit:

c(x + h) c(x)h

= average increase in cost per unit

The limit of the ratio as h

0 is the marginal value when x units

of commodity are produced:

c(x) = limh0

c(x + h) c(x)h

= marginal cost


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If the company, currently producing x units, increases itsproduction by one unit, then the additional cost c of producingthat one unit is

c = c(x + 1) c(x) c(x) 1 = c(x).

Thus,

Marginal cost estimates the cost of producing one unit beyond

the present production level.


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2.4.2 OTHER RATES OF CHANGE

Marginal Revenue and Marginal Profit Suppose r(x) is therevenue generated when x units of a particular commodity areproduced, and p(x) is the corresponding profit. When x = a unitsare being produced, then

The marginal revenue is r(a). It approximates r(a + 1) r(a),the additional revenue generated by producing one more unit.

The marginal profit is p(a). It approximates p(a + 1) p(a),the additional profit generated by producing one more unit.

Note Revenue = (number of units sold)(price per unit)


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Example 4.5 Marginal Cost. Suppose it costs

c(x) = x3 6x2 + 15xdollars to produce x stoves and your shop is currently producing 10stoves a day. About how much extra will it cost to produce one

more stove a day?

Solution The cost of producing one more stove a day when 10are produced is about c(10). Since

c(x) = (x3

6x2

+ 15x) = 3x2

12x + 15,c(10) = 3(100) 12(10) + 15 = 195.

Thus the additional cost will be about $195 if you produce 11stoves a day.


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Example 4.6 Marginal Revenue. If

r(x) = x3 3x2 + 12xgives the dollar revenue from selling x thousand candy bars, themarginal revenue when x thousand are sold is

r(x) = (x3 3x2 + 12x) = 3x2 6x + 12.The marginal revenue function estimates the increase in revenuethat will result from selling one additional unit. If you currently sell

10 thousand candy bars a week, you can expect your revenue toincrease by about

r(10) = 3(100) 6(10) + 12 = $252if you increase sales to 11 thousand bars a week.


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Example 4.7 Marginal Revenue and Profit. A manufacturerestimates that when x units of a particular commodity areproduced, the total cost will be c(x) = 18 x

2 + 3x + 98 dollars, andfurthermore, that all x units will be sold when the price isp(x) = 13 (75

x) dollars per units.

(a) Find the marginal cost and the marginal revenue.

(b) Use marginal cost to estimate the cost of producing the ninthunit.

(c) What is the actual cost of producing the ninth unit?

(d) Use marginal revenue to estimate the revenue derived fromthe sale of the ninth unit.

(e) What is the actual revenue from the sale of the ninth unit?


2.5 DERIVATIVE FTRIGONOMETRIC FUNCTIONS
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Derivative of the Sine

ddx

(sin x) = cos x

Derivative of the Cosine

ddx

(cos x) = sin x

Example 5.1 A body hanging from a spring is stretched 5 unitsbeyond its rest position and released at time t = 0 to bob up anddown. Its position at any latter time t is

s = 5 cos t.

What are its velocity and acceleration at time t?Dr. Nguyen Ngoc Hai CALCULUS I

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Because sin x and cos x are differentiable functions of x, therelated functions

tan x =sin x

cos xsec x =

1

cos x

cot x =cos x

sin xcsc x =

1

sin x

are differential functions. Their derivatives are

d

dx(tan x) =

1

cos2 x= sec2 x

d

dx(sec x) =

sin x

cos2 x= sec xtan x

ddx

(cot x) = 1sin2 x

ddx

(csc x) = csc xcot x

Example 5.2 Differentiate(a) y = x2 sin x (b) y = 3

xcot x.


2.6 HAIN RULE ANDIMPLICIT DIFFERENTIATION
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2.6.1 THE CHAIN RULE

Theorem 6.1 If g is differentiable at x and f is

differentiable at g(x), the composite function y = f g isdifferentiable at x and (f g)(x) is given by the product:

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


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In Leibniz notation,

If y = f(u) is a differentiable function of u and u = g(x) is adifferentiable function of x, then

dy

dx=

dy

du du

dx

ord

dxf(u) = f(u) du

dx


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Example 6.1 Find (f

g)(x) at x = 2 iff and g are

differentiable functions with g(2) = 3, g(2) = 4, and f(3) = 15.

Example 6.2 Find F(x) if F(x) = sin(x2 + 1)

Example 6.3 Find dy/dx at x = 0 ify = cos2 3x

.

Example 6.4 Find dy/dx for

y = 1 x

1 + x2

2.


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2.6.2 IMPLICIT DIFFERENTIATION

The functions that we have met so far can be described by

expressing one variable explicitly in terms of another variable. Afunction of this form is said to be in explicit form.

Sometimes practical problems will lead to equations in which thefunction y is not written explicitly in terms of the independent

variables x; for example, equations such as:

x2 + y2 = 25, (1)

x3 + y3 = 6xy. (2)

cos(ty) =t2

y (3)

Since it has not been solved for y, such an equation is said todefine y implicitly as a function of x and the function y is said tobe in implicit form.


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There is a simple technique based on the Chain Rule that youcan use to find dy/dx without first solving for y explicitly.

This technique is called implicit differentiation.

Example 6.5 Find an equation of the tangent line to the curvehaving equation

x2 + y2

3xy + 4 = 0

at the point (2, 4).



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IMPLICIT DIFFERENTIATION

Suppose an equation defines y implicitly as a differentiablefunction of x. To find dydx

:

1. Differentiate both sides of the equation with respect to x.

2. Solve the differentiated equation algebraically for dydx

.



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Example 6.6 Find dy/dt where cos(ty) = t2

y.

Example 6.7 Suppose that the output at an certain factory isQ = 2x3 + x2y + y3 units, where x is the number of hours ofskilled labor used and y is the number of hours of unskilled labor.The current labor force consists of 30 hours of skilled labor and 20hours of unskilled labor. Use calculus to estimate the change in

unskilled labor y that should be made to offset a 1-hour increase inskilled labor x so that output will be maintained at its current level.


2.7 DERIVATIVES OF INVERSE FUNCTIONS2.7.1 GENERAL FORMULA
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Theorem 7.1 Assume that f(x) is differentiable andone-to-one with inverse g(x) = f1(x). If a belongs to thedomain of g(x) and f

g(a)

= 0, then g(a) exists and

g(a) = 1f

g(a)

In Leibniz notation,

df

1

dx =1dfdx


2.7 DERIVATIVES OF INVERSE FUNCTIONS2.7.2 DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS
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For 1 < x < 1,

ddx

sin1 x = 11 x2

d

dxcos1 x = 1

1 x2

Example 7.1 Calculate f( 12 ), where f(x) = arcsin(x2).

d

dx tan1(x) =

1

1 + x2d

dx cot1(x) =

1

1 + x2

Example 7.2 Differentiate f(x) = xarctan

x.


2.7 DERIVATIVES OF INVERSE FUNCTIONS2.7.3 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
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dy

dx

(ex) = exdy

dx

(ax) = ax (a > 0)

dy

dx(ln |x|) = 1

x

dy

dx(loga |x|) =

1

x ln a(a > 0, a = 1)

Example 7.3 Find the derivative of

(a) f(x) = esin x (b) g(x) = 103x (c) h(x) = log2

x2 + 1.

If we combine the above formulae with the Chain Rule, we get

dy

dx(eu) = eu

du

dx

dy

dx(au) = au

du

dxdy

dx(ln |u|) = 1

u

du

dx

dy

dx(loga |u|) =

1

uln a

du

dx


2.7 DERIVATIVES OF INVERSE FUNCTIONS2.7.4 LOGARITHMIC DIFFERENTIATION

Th l l i f d i i f li d f i i l i
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The calculation of derivatives of complicated functions involvingproducts, quotients, or powers can often be simplified by taking

logarithms. The method used in the following examples is calledlogarithmic differentiation.


f(x) = (sin x)x, 0 < x < .

In general, to differentiate a function of the form

y = u(x)v(x)

we take natural logarithms of both sides and differentiate implicitly.


y =(x + 1)2(2x2 3)

x2 + 1.


2.8 LINEAR APPR XIMATI NAND APPLICATIONS

2 8 1 LINEAR APPROXIMATION
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2.8.1 LINEAR APPROXIMATION

Definition 8.1 The approximation

f(x) f(a) + f(a)(x a) (4)

is called the linear approximation or tangent lineapproximation of f at x = a, and the function

L(x) = f(a) + f(a)(x a)

is called the linearization of f at x = a.

Example 8.1 Find the linearization of the function f(x) =

x

at a = 1 and use it to approximate the number

1.001.



2 8 1 LINEAR APPROXIMATION
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Let x = x

a be the change in x and y = f(x)

f(a) the

corresponding change in y. Then (4) can be written as

y f(a)x.The right side f(a)x is called the differential of f at a anddenoted by df or dy:

dy = df = f(a)x (5)

In particular, if f(x) = x, then df = dx = x at every point a.Substituting dx = x into Equation (5) we get

dy = df = f(a)dx (6)We call dx the differential of x, and df the correspondingdifferential of y. If we divide both sides of the Equation (6) bydx, we obtain the familiar equation

df = Dr. Nguyen Ngoc Hai CALCULUS I


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Let P(x, f(x)) and Q(x + x, f(x + x)) be points on the graph

of f and let dx = x. The corresponding change in y is

y = f(x + x) f(x).

Therefore,

dy represents the amount that the tangent line rises or fall,whereas y represents the amount that the curve y = f(x)rises or falls when x changes an amount dx .

Sincey = f(x + x) f(x) dy,

we havef(x + x) f(x) + dy.



2.8.2 DIFFERENTIALS
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Example 8.2 The radius of a circle increases from an initialvalue of r0 = 10 by an amount dr = 0.1. Estimate thecorresponding increase in the circles area A = r2 by calculatingdA. Compare dA with the true change A.

The percentage error is defined by

percentage error =

error

actual value

100%

The percentage error is often more important than the error itself.



2.8.3 THE SIZE OF THE ERROR
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In any approximation, the error is defined by

error =true value approximate value

If the linearization of y = f(x) about x = a,

L(x) = f(a) + f(a)(x

a),

is used to approximate f(x) near a, then the error E(x) in thisapproximation is

E(x) = |f(x) L(x)| = |f(x) f(a) f(a)(x a)| = |y dy|.

Theorem 8.1 (Error Bound) If|f(x)| K for all x inthe interval between a and a + h, then

E(x) 12

Kh2.

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Pages Exercises Assignments

148-150 6, 8 11, 13, 16, 22, 24, 25155-156 3, 5, 8, 21, 35 6, 22, 28, 31, 36167-170 3, 21, 31 11, 30, 32, 41, 43197-199 14, 17, 42, 63 18, 20, 34, 41, 45,

47, 54, 55, 60, 61204-205 9, 17, 27, 40, 4, 12, 15, 20, 21,28, 30, 32, 34, 38

215-218 1, 3, 6, 2, 4, 5, 7, 10, 11, 14,15, 16, 19, 22,23


P E i A i
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Pages Exercises Assignments

223-225 4, 12, 20, 24, 31, 11, 17, 25, 26, 29,

33, 37 32, 34233-236 10, 16, 33, 42, 19, 20, 22, 30, 32,

45, 54 34, 37, 39, 40,46, 47, 48, 49, 5558, 60, 71, 74

243-245 8, 10, 13, 17, 6, 11, 12, 14, 16,20, 26, 29, 46 18, 19, 23, 24, 32,

38, 45, 47, 48, 53250-251 4, 6, 12, 28 11, 16, 19, 20, 23,

29, 33, 35, 38, 39256-257 3, 6, 8 16, 18, 19

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