Technique of Differentiation Ppt 03

8/12/2019 Technique of Differentiation Ppt 03

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2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12eSlide 1 of 33

Chapter 3

Techniques of Differentiation


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The Product and Quotient Rules

The Chain Rule and the General Power Rule

Implicit Differentiation and Related Rates

Chapter Outline


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3.1The Product and Quotient Rules


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The Product Rule

The Quotient Rule

Rate of Change

Section Outline


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The Product Rule


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The Product Rule

EX MPLE

SOLUTION

Differentiate the function.

Let and . Then, using the product rule, and thegeneral power rule to compute g(x),

1022 33 xx

32

xxf

102

3

xxg

333333 210210221022 xdx

dxx

dx

dxxx

dx

d

xxxdxd

xx 2333103 1022922

.2323103 102922 xxxxx


7/33 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12eSlide 7 of 33

The Quotient Rule



The Quotient Rule

EX MPLE

SOLUTION

Differentiate.

Let and . Then, using the quotient rule

x

xx 34 24

34 24

xxxf xxg

2

242424 3434

34

x

xdx

dxxxx

dx

dx

x

xx

dx

d

2

243 13484

x

xxxxx

22

22

2

2

4

2

24

343343343

xx

xx

x

x

x

x

xx

Now simplify.



The Quotient Rule

Now lets differentiate again, but first simplify the expression.

xx

x

x

x

x

xx 3434 2424

Now we can differentiate the function in its new form.

CONTINUED

13 34 xxx

2213 34334 xxxxxdx

d

Notice that the same answer was acquired both ways.



Rate of Change

EX MPLE

SOLUTION

(Rate of Change) The width of a rectangle is increasing at a rate of 3 inches

per second and its length is increasing at the rate of 4 inches per second. At

what rate is the area of the rectangle increasing when its width is 5 inches and

its length is 6 inches? [Hint: Let W(t) andL(t) be the widths and lengths,

respectively, at time t.]

Since we are looking for the rateat which the area of the rectangle is changing,

we will need to evaluate the derivative of an area function,A(x) for those given

values (and to simplify, lets say that this is happening at time t= t0). Thus

tWtLtA This is the area function.

tLtWtWtLtA Differentiate using the productrule.


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The Product Rule & Quotient Rule

Another way to order terms in the product and quotient rules, for the purpose ofmemorizing them more easily, is

xfxgxgxfxgxfdx

d

PRODUCT RULE

.2

xg

xfxgxgxf

xg

xf

dx

d

QUOTIENT RULE



3.2The Chain Rule and the General Power Rule



The Chain Rule

Marginal Cost and Time Rate of Change

Section Outline



The Chain Rule


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The Chain Rule

EX MPLE

SOLUTION

Use the chain rule to compute the derivative of f(g(x)), where

and .

Finally, by the chain rule,

24 xx

xf 41 xxg

,2

4

2 x

xxf

34xxg

424

121

4x

xxgf

.4121

4 3424

xxx

xgxgfxgfdx

d


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The Chain Rule

EX MPLE

SOLUTION

Compute using the chain rule.

Sinceyis not given directly as a function ofx, we cannot compute bydifferentiating ydirectly with respect tox. We can, however, differentiate with

respect to uthe relation , and get

dxdy

22,1 xuuy

dx

dy

1 uy

.12

1

udu

dy

Similarly, we can differentiate with respect to xthe relation and get22xu

.4xdx

du


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The Chain Rule

Applying the chain rule, we obtain

.412

1x

udx

du

du

dy

dx

dy

It is usually desirable to express as a function ofxalone, so we substitute

2x2 for uto obtain

CONTINUED

dx

dy

.

122

4

2

x

x

dx

dy


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Marginal Cost & Time Rate of Change

EX MPLE

SOLUTION

(Marginal Cost and Time Rate of Change) The cost of manufacturingxcases of

cereal is Cdollars, where . Weekly production at tweeks from

the present is estimated to bex= 6200 + 100tcases.

.dx

dC

243 xxC

(a) Find the marginal cost,

(b) Find the time rate of change of cost,

(c) How fast (with respect to time) are costs rising when t= 2?

.dt

dC

x

xxdx

d

dx

dC 23243

(a) We differentiate C(x).


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Marginal Cost & Time Rate of Change

(b) To determine , we use the Chain Rule.dt

dC

1001006200,2

3 tdt

d

dt

dx

xdx

dC

Now we rewritexin terms of tusingx= 6200 + 100t.

CONTINUED

10023

xdt

dx

dx

dC

dt

dC

10010062002

3

tdt

dC

(c) With respect to time, when t= 2, costs are rising at a rate of

.per weekcerealofcases5.302100

21006200

23

2

tdt

dC


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3.3Implicit Differentiation and Related Rates


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Implicit Differentiation


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EX MPLE

SOLUTION

Use implicit differentiation to determine the slope of the graph at the given

point.1,3;54 23 yxxy

The second term,x2, has derivative 2xas usual. We think of the first term, 4y3,as having the form 4[g(x)]3. To differentiate we use the chain rule:

xgxgxgdx

d

23124

or, equivalently,

.124 23dx

dyyy

dx

d


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On the right side of the original equation, the derivative of the constant

function -5 is zero. Thus implicit differentiation of yields

.612

222

y

x

y

x

dx

dy

Solving for we have

.0212 2 xdx

dyy

CONTINUED

54 23 xy

dxdy

At the point (3, 1) the slope is

.2

1

6

3

16

3

6 2

1

32

1

3

y

x

y

x y

x

dx

dy


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This is the general power rule for implicit differentiation.


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Related Rates


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Related Rates

EX MPLE(Related Rates) An airplane flying 390 feet per second at an altitude of 5000feet flew directly over an observer. The figure below shows the relationship of

the airplane to the observer at a later time.

(a) Find an equation relatingxandy.

(b) Find the value ofxwhenyis 13,000.

(c) How fast is the distance from the observer to the airplane changing at the

time when the airplane is 13,000 feet from the observer? That is, what is

at the time when andy= 13,000?

.

dx

dy

390dt

dx


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Related Rates

SOLUTION(a) To find an equation relatingxandy, we notice thatxandyare the lengths of

two sides of a right triangle. Therefore

2225000 yx

CONTINUED

.000,000,25 22

yx

(b) To find the value ofxwhenyis 13,000, replaceywith 13,000.

22000,000,25 yx This is the function from part (a).

22

000,13000,000,25 x Replaceywith 13,000.

000,000,169000,000,25 2 x Square.

000,000,1442 x Subtract.


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l d


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Related Rates

CONTINUED

Therefore, the rate at which the distance from the plane to the observer is

changing for the given values is 360 ft/sec.

dt

dy000,132390000,122 y= 13,000;x= 12,000; .390

dt

dx

dt

dy000,26000,360,9 Simplify.

dtdy360 Divide.

Technique of Differentiation Ppt 03

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