chapter 13 10-13-09kappraff/chapter13.pdf · chapter 13 product, qutient, chain rule and trig ......

Calculus and Structures 192


PRODUCT, QUOTIENT, CHAIN RULE, AND TRIG FUNCTIONS

CHAPTER 13

Copyright

Calculus and Structures

Chapter 13 PRODUCT, QUTIENT, CHAIN RULE AND TRIG FUNTIONS

194

13.1 NEW FUNCTIONS FROM OLD ONES Given two functions f(x) and g(x) we can define three new functions in terms of these old ones:

(fg)(x) = f(x)g(x) ; )()())((

xgxfx

gf

; and ))(())(( xgfxgf (see Section 2.6. These new

functions are called the product, quotient, composition of f and g. If the derivatives of f and g,

f’(x) and g’(x), are known, then we can compute the derivatives of fg, gf , and gf in terms of

f’(x) and g’(x).

13.2 THE DERIVATIVE OF THE PRODUCT OF TWO FUNCTIONS To find the derivative of (fg)(x) use the derivative machine:

)()()()()()()'( 00 homhxgxfxgxfxfg (1) where, )()(')()( 00 hohxfxfxf (2a) )()(')()( 00 hohxgxgxg (2b)

To solve for m, insert Eq. 2a and b in Eq. 1 to get,

))(()('()((()()())(( 100 hohxfxfxgxfxfg ))()(')( 200 hohxgxg hxgxfxgxfxgxf ))()(')(')((()()( 000000 (little o(h) terms) (3) Eliminate the o(h) terms as usual and, after some algebra, )()(')(')()()'( xgxfxgxfmxfg (4) Eq. 4 is known as the product rule. Example:1 Use the product rule to find the derivative of (fg)(x) = )1)(2( 32 xxx . where

xxxf 2)( 2 , 1)( 3 xxg and , 22)(' xxf , 23)(' xxg . .From Eq. 4, )1)(22()3)(2()()'( 322 xxxxxxfg

Example 2

195Calculus and Structures

Section 13.2

Find . dx

exd x )( 2

.

Let f(x) = x2 and g(x) = ex where f’(x) = 2x and g’(x) = ex . From Eq. 4,

xxx

xeexdx

exd 2)( 22

Problems: a) ) Find: dy/dx3)(52y= ( 22 xxx

b) We showed in Sec. 9.6 that xdx

xd sin)(cos . Use this with the help of the product rule

to compute: dx

xxd )cos(

13.3 THE DERIVATIVE OF THE QUOTIENT OF TWO FUNCTIONS The following rule enables you to compute the derivative of the quotient of two functions if you know the derivatives of the two functions.. The proof follows in a similar manner to the Product Rule shown above. I leave the details to the student.

2))(()'()()'()()'

)()(()()'(

xgxgxfxfxg

xgxfx

gf

(5)

Eq. 5 is known as the quotient rule. Example 3

Find, dx

xxd )

1( 2

2

.

Let f(x) = x2 and g(x) = x2 – 1 where f ‘(x) = 2x and g ‘(x) = 2x. Replacing these functions in Eq. 5,

22

222

2

)1()(2)2)(1()

1(

x

xxxxdx

xxd

22 )1(-2

x

x



13.4 THE DERIVATIVE OF A FUNCTION OF A FUNCTION – THE CHAIN RULE. Consider ))(( xgf f(g(x)). I will show by using the Derivative machine that its derivative at x0 is: )())((')()'( 00 xgxgfxgf Using the Derivative machine: )()(')()( 1000 hohxgxghxg )()(')()( 2000 hohxfxfhxf ))()'()(())(())(( 20000 hohxgxgfhxgfhxgf Set o2(h) to zero and let khxg )(' 0 ))(())(( 00 kxgfhxgf = kxgfxgf ))(('))(( 00 )()('))(('))(())(( 0000 hohxgxgfxgfhxgf Therefore, ‘ )())((')()'( 00 xgxgfxgf (6) Example 4

Find dx

xd 21

Let 2/1)( xxf and g(x) = x2 + 1, 2/1

21)(' xxf and g’(x) = 2x..

From Eq. 6,

)2()1(211 2/12

2

xxdx

xd

=21 x

x

Successfully finding the derivative of the function of a function often presents the student with considerable difficulty. There is another approach that I call the “box method” which often simplifies this computation. We illustrate this for the case of Example 4. One first


Section 13.4

sketches a diagram of the compound function by representing it as a box with an outside function and an inside function as shown in Fig. 1. The function is gotten by placing the inside function, x2 + 1, into the open parenthesis of the outside function, ( )1/2

In order to differentiate the compound function, we use the following step by step procedure:

a) Begin with the outside function. Its derivative is: 21 ( )-1/2 .; b) replace the open

parenthesis by whatever is in the box, i.e., 21 (x2 + 1)-1/2 ; c) Differentiate the inside

function, i.e., 2x; d) multiply the result of b) and c) to get, )2()1(211 2/12

2

xxdx

xd .

Example 5 Sometimes you need several boxes in order to define the function . Use the “box method” to find,

dx

xd 3)3(cos

First sketch the function. We need three boxes to do this. The outermost function is the square function, ( )2 , the next function is the cosine function, i.e., cos ( ), while the innermost function is the 3x function (see Fig. 2).

Using a slightly enhanced version of the procedure, a) d ( )3/dx = 3 ( ) ; b) Insert everything inside the outer box into the open parenthesis, i.e.,

2))3(cos(3 x ; c) d cos ( ) /dx = -sin ( ); d) Insert everything within the second box into the open parenthesis, i.e., - sin (3x); e) differentiate the innermost function, i.e., d (3x)/dx = 3 ; f) multiply the results of b, d, and e to get,

Fig. 1

Fig.2

2

cos( ).( ) 3

x3

( ) 1/2

1+x2



)3sin()3)(cos3(3)3(cos 23

xxdx

xd

)3sin()3(cos9 2 xx 13.5 THE CHAIN RULE Computation of the derivative of compound functions is often referred to as the chain rule. Why do we use this terminology? Let us illustrate the chain rule for Example 4. We can define the compound function by a “chain” of functions. Example 6 Define 2/12 )1( xy as follows, 2/1)(uy 12 xu Therefore we have a chain of dependencies, yux In other words, knowing x you can find u and then knowing u you can find y. To find the derivative of y with respect to x just differentiate the function down the chain as follows:

)2(21))(( 2/1 xu

dxdu

dudy

dxdy

= 21 x

x

where we have inserted the value of u in terms of x. Remark 1: The chain rule merely makes the box method more formal. Let us apply the chain rule to Example 5 to find the derivative of . Example 7 Define )3(cos3 xy as, 3)(uy

cos 3x )(3


Section 13.5

u = cos (w) w = 3x The chain for this function is:

yuwx where,

)3)(sin(3))()(( 2 wududw

dwdu

dudy

dxdy

)3)(sin(cos3 2 ww )3sin()3(cos9 2 xx

where u is expressed in terms of w and then w is expressed in terms of x.

13.6 THE INVERSE OF THE CHAIN RULE – THE SUBSTITUTION METHOD OF FINDING ANTIDERIVATIVES

In Sections 13.4 and 13.5 we have computed the derivative of compound functions. In this section we reverse the chain rule and find antiderivatives of certain compound functions using the substitution method. For example,. Example 8 Find, dxx 2/1)31( (7a) We know how to find., dxx 2/1 , (7b) and the following procedure that will reduce Expression 7a to 7b. Let’s see how it works. a) Let u = 1 + 3x (8)

b) Define, dxdxdudu , or du = 3 dx

c) Solve for dx, i.e., dudx31

d) Replace the result of a) and c) into Eq. 7a,



duu 2/1)(31

d) Take the constant out of the antiderivate sign using the property b) in Section 10.5, i.e.,

duu 2/1

31 (9)

e) Use the power law (property a) of Section 10.5 to anti differentiate Eq. 9,

cu 2/3)32(

31 (10)

f) Express u in Expression 10 by x using Eq. 8,

cxdxx 2/12/1 )31(92)31( .

Remark 2: This substitution method will work whenever the derivative of the substitution, is sitting under

the integral sign differing only by a constant, e.g., in Example 8, 3)31(

dx

xd . In other

words, the function to be antidifferentiated is off by the constant 3. Example 9 Use this procedure to compute, dxxe x

2

(11)

a) Let u = x2

b) dxdxdudu or du = 2x dx

c) dux

dx21

d) duex

x u2

21 or dueu2

1 (cancelling x)

e) ceu 21

f) cedxxe xx 22

21


Section 13.6

Remark 3: Since xdx

dx 22

and there is an x under the antiderivative sign in expression 11, so

the function to be antidifferentiated differs only by the constant, 2. Therefore the method will work. Remark 4: We cannot use the substitution method to find the antiderivative of dxe x

2

. (why?) 13.7 TRIG FUNCTIONS The sine and cosine functions of trigonometry are defined in terms of a unit circle. For the unit circle in Fig. 3, if we consider the ray coming from the origin in the direction of the positive x-axis as 0 deg. Then )sin,(cos is the point on the unit circle at a positive angle of deg. measured from the x-axis in a counterclockwise direction, and negative angle in the clockwise direction. You can see from Fig. 3, using the Pythagorean Theorem, that 1cossin 22 . (12)

In Section 9. we found that,

sincos

dxd

We can now use this to find the derivative of .sin From the right triangle in Fig. 4 we see that,

sin)2

cos( , (13)

cos)2

sin( . (14)

Fig.3

( cos , sin )θ θ

cos

sin

1

1



Fig.4Example 10

Show that, cossin

dxxd .

Using Eq. 13,

dx

xd

dxxd )

2cos(sin

Using the chain rule,

)2

sin()1(sin xdx

xd

xcos Example 11:

Show that xdx

xd 2sectan

Let xxx

cossintan .

Using the quotient rule,

)cossin(tan

xx

dxd

dxxd

x

xxxxdx

xd2cos

sin)sin()(coscostan

Using Eq. 12,

xxdx

xd 22 sec

cos1tan

2c

a

b


Section 13.6

Example 12

Show that 1

1tan2

1

xdxxd where x1tan is the inverse tangent function (see Sec. 2.1 and

2.7) In Section 2.7 we saw that if )(xfy then ))(()( 11 xffyf = x, i.e., )(1 yfx Applying this to xy 1tan we have that yx tan In order to find the derivative of an inverse function we state without proof that,

dydxdx

dy/1

(15)

or,

dyyddx

xd/tan

1tan 1

(16)

From Example 11,

ydy

yd 2sectan (17)

Replacing Eq. 17 in 16,

ydx

xd2

1

sec1tan

But, xy 1tan so that we have that,

21

1

))(sec(tan1tan

xdxxd

(18)

where )sec(tan 1 x means “the secant of the angle whose tangent is x.” Let’s call this angle referring to Fig 5,



We see that 1)sec(tan 21 xx And replacing this in Eq. 18 gives,

1

1tan2

1

xdxxd

Once we have a new derivative we also have a new antiderivative so that,

cxdxx

12 tan)

11(

Fig.51

x

12 x


Chapter 13 Problems

Problems

1


Chapter 13 Problems

chapter 13 10-13-09kappraff/chapter13.pdf · chapter 13 product, qutient, chain rule and trig ......

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