266
DIFFERENTIATION
CHAPTER 3
57. Let f .x/ D x sin x and g.x/ D x cos x.
(a) Show that f 0 .x/ D g.x/ C sin x and g 0 .x/ D !f .x/ C cos x.
(b) Verify that f 00 .x/ D !f .x/ C 2 cos x and
g 00 .x/ D !g.x/ ! 2 sin x.
(c) By further experimentation, try to find formulas for all higher derivatives of f and g. Hint: The kth derivative depends on
whether k D 4n, 4n C 1, 4n C 2, or 4n C 3.
Let f .x/ D x sin x and g.x/ D x cos x.
(a) We examine first derivatives: f 0 .x/ D x cos x C .sin x/ " 1 D g.x/ C sin x and g 0 .x/ D .x/.! sin x/ C .cos x/ " 1 D
!f .x/ C cos x; i.e., f 0 .x/ D g.x/ C sin x and g 0 .x/ D !f .x/ C cos x.
(b) Now look at second derivatives: f 00 .x/ D g 0 .x/ C cos x D !f .x/ C 2 cos x and g 00 .x/ D !f 0 .x/ ! sin x D !g.x/ ! 2 sin x;
i.e., f 00 .x/ D !f .x/ C 2 cos x and g 00 .x/ D !g.x/ ! 2 sin x.
! The third derivatives are f 000 .x/ D !f 0 .x/ ! 2 sin x D !g.x/ ! 3 sin x and g 000 .x/ D !g 0 .x/ ! 2 cos x D f .x/ !
(c)
3 cos x; i.e., f 000 .x/ D !g.x/ ! 3 sin x and g 000 .x/ D f .x/ ! 3 cos x.
! The fourth derivatives are f .4/ .x/ D !g 0 .x/ ! 3 cos x D f .x/ ! 4 cos x and g .4/ .x/ D f 0 .x/ C 3 sin x D g.x/ C 4 sin x;
i.e., f .4/ D f .x/ ! 4 cos x and g .4/ .x/ D g.x/ C 4 sin x.
! We can now see the pattern for the derivatives, which are summarized in the following table. Here n D 0; 1; 2; : : :
SOLUTION
k
f
.k/ .x/
g .k/ .x/
4n
4n C 1
4n C 2
4n C 3
f .x/ ! k cos x
g.x/ C k sin x
!f .x/ C k cos x
!g.x/ ! k sin x
g.x/ C k sin x
!f .x/ C k cos x
!g.x/ ! k sin x
f .x/ ! k cos x
58.
Figure 2 shows the geometry behind the derivative formula .sin !/0 D cos !. Segments BA and BD are parallel to the
x- and y-axes. Let " sin ! D sin.! C h/ ! sin !. Verify the following statements.
(a) " sin ! D BC
(b) †BDA D ! Hint: OA ? AD.
(c) BD D .cos !/AD
Now explain the following intuitive argument: If h is small, then BC # BD and AD # h, so " sin ! # .cos !/h and .sin !/0 D
cos !.
y
D
C
A
B
h
θ
x
O
1
FIGURE 2
SOLUTION
(a) We note that sin.! C h/ is the y-coordinate of the point C and sin ! is the y-coordinate of the point A, and therefore also of
the point B. Now, " sin ! D sin.! C h/ ! sin ! can be interpreted as the difference between the y-coordinates of the points B and
C ; that is, " sin ! D BC .
(b) From the figure, we note that †OAB D !, so †BAD D # ! ! and †BDA D !.
(c) Using part (b), it follows that
cos ! D
BD
AD
or
BD D .cos !/AD:
For h “small,” BC # BD and AD is roughly the length of the arc subtended from A to C ; that is, AD # 1.h/ D h. Thus, using
(a) and (c),
" sin ! D BC # BD D .cos !/AD # .cos !/h:
In the limit as h ! 0,
" sin !
! .sin !/0 ;
h
so .sin !/0 D cos !.
S E C T I O N 3.7
The Chain Rule
267
3.7 The Chain Rule
Preliminary Questions
1. Identify the outside and inside functions for each of these composite functions.
p
(a) y D 4x C 9x 2
(b) y D tan.x 2 C 1/
(c) y D sec5 x
(d) y D .1 C e x /4
SOLUTION
(a)
(b)
(c)
(d)
p
The outer function is x, and the inner function is 4x C 9x 2 .
The outer function is tan x, and the inner function is x 2 C 1.
The outer function is x 5 , and the inner function is sec x.
The outer function is x 4 , and the inner function is 1 C e x .
2. Which of the following can be differentiated easily without using the Chain Rule?
x
(a) y D tan.7x 2 C 2/
(b) y D
xC1
p
p
(c) y D x " sec x
(d) y D x cos x
(e) y D xe x
(f) y D e sin x
p
x
The function xC1
can be differentiated using the Quotient Rule, and the functions x " sec x and xe x can be
p
differentiated using the Product Rule. The functions tan.7x 2 C 2/, x cos x and e sin x require the Chain Rule.
SOLUTION
3. Which is the derivative of f .5x/?
(a) 5f 0 .x/
SOLUTION
(b) 5f 0 .5x/
(c) f 0 .5x/
The correct answer is (b): 5f 0 .5x/.
4. Suppose that f 0 .4/ D g.4/ D g 0 .4/ D 1. Do we have enough information to compute F 0 .4/, where F .x/ D f .g.x//? If not,
what is missing?
If F .x/ D f .g.x//, then F 0 .x/ D f 0 .g.x//g 0 .x/ and F 0 .4/ D f 0 .g.4//g 0 .4/. Thus, we do not have enough
information to compute F 0 .4/. We are missing the value of f 0 .1/.
SOLUTION
Exercises
In Exercises 1–4, fill in a table of the following type:
f .g.x//
f 0 .u/
f 0 .g.x//
g 0 .x/
.f ı g/0
.f ı g/0
1. f .u/ D u3=2 , g.x/ D x 4 C 1
SOLUTION
f .g.x//
f 0 .u/
f 0 .g.x//
g 0 .x/
.x 4 C 1/3=2
3 1=2
2u
3 4
2 .x
4x 3
C 1/1=2
6x 3 .x 4 C 1/1=2
2. f .u/ D u3 , g.x/ D 3x C 5
SOLUTION
f .g.x//
f 0 .u/
f 0 .g.x//
g 0 .x/
.3x C 5/3
3u2
3.3x C 5/2
3
f .g.x//
f 0 .u/
f 0 .g.x//
g 0 .x/
tan.x 4 /
sec2 u
sec2 .x 4 /
4x 3
.f ı g/0
9.3x C 5/2
3. f .u/ D tan u, g.x/ D x 4
SOLUTION
4. f .u/ D u4 C u, g.x/ D cos x
.f ı g/0
4x 3 sec2 .x 4 /
268
DIFFERENTIATION
CHAPTER 3
SOLUTION
f .g.x//
f 0 .u/
f 0 .g.x//
g 0 .x/
.cos x/4 C cos x
4u3 C 1
4.cos x/3 C 1
! sin x
.f ı g/0
!4 sin x cos3 x ! sin x
In Exercises 5 and 6, write the function as a composite f .g.x// and compute the derivative using the Chain Rule.
5. y D .x C sin x/4
SOLUTION
Let f .x/ D x 4 , g.x/ D x C sin x, and y D f .g.x// D .x C sin x/4 . Then
dy
D f 0 .g.x//g 0.x/ D 4.x C sin x/3 .1 C cos x/:
dx
6. y D cos.x 3 /
SOLUTION
Let f .x/ D cos x, g.x/ D x 3 , and y D f .g.x// D cos.x 3 /. Then
dy
D f 0 .g.x//g 0.x/ D !3x 2 sin.x 3 /:
dx
d
cos u for the following choices of u.x/:
dx
(a) u D 9 ! x 2
(b) u D x "1
7. Calculate
(c) u D tan x
SOLUTION
(a) cos.u.x// D cos.9 ! x 2 /.
d
cos.u.x// D ! sin.u.x//u0 .x/ D ! sin.9 ! x 2 /.!2x/ D 2x sin.9 ! x 2 /:
dx
(b) cos.u.x// D cos.x "1 /.
!
"
d
1
sin.x "1 /
cos.u.x// D ! sin.u.x//u0.x/ D ! sin.x "1 / ! 2 D
:
dx
x
x2
(c) cos.u.x// D cos.tan x/.
d
cos.u.x// D ! sin.u.x//u0 .x/ D ! sin.tan x/.sec2 x/ D ! sec2 x sin.tan x/:
dx
d
f .x 2 C 1/ for the following choices of f .u/:
dx
(a) f .u/ D sin u
(b) f .u/ D 3u3=2
8. Calculate
(c) f .u/ D u2 ! u
SOLUTION
(a) Let sin.u/ D sin.x 2 C 1/. Then
d
d 2
sin.x 2 C 1/ D cos.x 2 C 1/ "
.x C 1/ D cos.x 2 C 1/2x D 2x cos.x 2 C 1/:
dx
dx
(b) Let 3u3=2 D 3.x 2 C 1/3=2 : Then
d
3
d 2
9
3.x 2 C 1/3=2 D 3 " .x 2 C 1/1=2
.x C 1/ D .x 2 C 1/1=2 .2x/ D 9x.x 2 C 1/1=2 :
dx
2
dx
2
(c) Let u2 ! u D .x 2 C 1/2 ! .x 2 C 1/: Then
$
d # 2
d
.x C 1/2 ! .x 2 C 1/ D Œ2.x 2 C 1/ ! 1$ .x 2 C 1/ D Œ2.x 2 C 1/ ! 1$.2x/ D 4x 3 C 2x:
dx
dx
df
df
du
9. Compute
if
D 2 and
D 6.
dx du
dx
SOLUTION
Assuming f is a function of u, which is in turn a function of x,
df
df du
D
"
D 2.6/ D 12:
dx
du dx
10. Compute
df ˇˇ
if f .u/ D u2 , u.2/ D !5, and u0 .2/ D !5.
ˇ
dx xD2
S E C T I O N 3.7
The Chain Rule
Because f .u/ D u2 , it follows that f 0 .u/ D 2u. Therefore,
ˇ
df ˇˇ
D f 0 .u.2//u0 .2/ D 2u.2/u0 .2/ D 2.!5/.!5/ D 50:
dx ˇxD2
SOLUTION
In Exercises 11–22, use the General Power Rule or the Shifting and Scaling Rule to compute the derivative.
11. y D .x 4 C 5/3
Using the General Power Rule,
SOLUTION
d 4
d 4
.x C 5/3 D 3.x 4 C 5/2
.x C 5/ D 3.x 4 C 5/2 .4x 3 / D 12x 3 .x 4 C 5/2 :
dx
dx
12. y D .8x 4 C 5/3
Using the General Power Rule,
SOLUTION
13. y D
p
SOLUTION
d
d
.8x 4 C 5/3 D 3.8x 4 C 5/2
.8x 4 C 5/ D 3.8x 4 C 5/2 .32x 3 / D 96x 3 .8x 4 C 5/2 :
dx
dx
7x ! 3
Using the Shifting and Scaling Rule
14. y D .4 ! 2x ! 3x 2 /5
SOLUTION
d p
d
1
7
7x ! 3 D
.7x ! 3/1=2 D .7x ! 3/"1=2 .7/ D p
:
dx
dx
2
2 7x ! 3
Using the General Power Rule,
d
d
.4 ! 2x ! 3x 2 /5 D 5.4 ! 2x ! 3x 2 /4
.4 ! 2x ! 3x 2 / D 5.4 ! 2x ! 3x 2 /4 .!2 ! 6x/
dx
dx
D !10.1 C 3x/.4 ! 2x ! 3x 2 /4 :
15. y D .x 2 C 9x/"2
SOLUTION
Using the General Power Rule,
d 2
d 2
.x C 9x/"2 D !2.x 2 C 9x/"3
.x C 9x/ D !2.x 2 C 9x/"3 .2x C 9/:
dx
dx
16. y D .x 3 C 3x C 9/"4=3
SOLUTION
Using the General Power Rule,
d 3
4
d 3
4
.x C 3x C 9/"4=3 D ! .x 3 C 3x C 9/"7=3
.x C 3x C 9/ D ! .x 3 C 3x C 9/"7=3 .3x 2 C 3/
dx
3
dx
3
D !4.x 2 C 1/.x 3 C 3x C 9/"7=3 :
17. y D cos4 !
SOLUTION
Using the General Power Rule,
d
d
cos4 ! D 4 cos3 !
cos ! D !4 cos3 ! sin !:
d!
d!
18. y D cos.9! C 41/
SOLUTION
Using the Shifting and Scaling Rule
d
cos.9! C 41/ D !9 sin.9! C 41/:
d!
19. y D .2 cos ! C 5 sin !/9
SOLUTION
Using the General Power Rule,
d
d
.2 cos ! C 5 sin !/9 D 9.2 cos ! C 5 sin !/8
.2 cos ! C 5 sin !/ D 9.2 cos ! C 5 sin !/8 .5 cos ! ! 2 sin !/:
d!
d!
p
20. y D 9 C x C sin x
269
270
DIFFERENTIATION
CHAPTER 3
SOLUTION
Using the General Power Rule,
21. y D e x"12
SOLUTION
d p
1
d
1 C cos x
9 C x C sin x D .9 C x C sin x/"1=2
.9 C x C sin x/ D p
:
dx
2
dx
2 9 C x C sin x
Using the Shifting and Scaling Rule,
d x"12
e
D .1/e x"12 D e x"12 :
dx
22. y D e 8xC9
SOLUTION
Using the Shifting and Scaling Rule,
d 8xC9
e
D 8e 8xC9 :
dx
In Exercises 23–26, compute the derivative of f ı g.
23. f .u/ D sin u, g.x/ D 2x C 1
SOLUTION
Let h.x/ D f .g.x// D sin.2x C 1/. Then, applying the shifting and scaling rule, h0 .x/ D 2 cos.2x C 1/. Alternately,
24. f .u/ D 2u C 1,
SOLUTION
d
f .g.x// D f 0 .g.x//g 0.x/ D cos.2x C 1/ " 2 D 2 cos.2x C 1/:
dx
g.x/ D sin x
Let h.x/ D f .g.x// D 2.sin x/ C 1. Then h0 .x/ D 2 cos x. Alternately,
d
f .g.x// D f 0 .g.x//g 0.x/ D 2 cos x:
dx
25. f .u/ D e u ,
SOLUTION
g.x/ D x C x "1
Let h.x/ D f .g.x// D e xCx
!1
. Then
#
$
d
!1
f .g.x// D f 0 .g.x//g 0 .x/ D e xCx
1 ! x "2 :
dx
u
, g.x/ D csc x
u!1
SOLUTION Let h.x/ D f .g.x//: Then, applying the quotient rule:
26. f .u/ D
h0 .x/ D
.csc x ! 1/.! csc x cot x/ ! .csc x/.! csc x cot x/
csc x cot x
D
:
2
.csc x ! 1/
.csc x ! 1/2
Alternately,
d
1
csc x cot x
f .g.x// D f 0 .g.x//g 0.x/ D !
.! csc x cot x/ D
;
dx
.csc x ! 1/2
.csc x ! 1/2
1
where we have used the quotient rule to calculate f 0 .u/ D ! .u"1/
2.
In Exercises 27 and 28, find the derivatives of f .g.x// and g.f .x//.
27. f .u/ D cos u, u D g.x/ D x 2 C 1
SOLUTION
d
f .g.x// D f 0 .g.x//g 0.x/ D ! sin.x 2 C 1/.2x/ D !2x sin.x 2 C 1/:
dx
d
g.f .x// D g 0 .f .x//f 0 .x/ D 2.cos x/.! sin x/ D !2 sin x cos x:
dx
1
28. f .u/ D u3 , u D g.x/ D
xC1
1
SOLUTION The derivative of xC1 is taken using the shifting and scaling rule.
!
"
"2 !
1
3
d
1
D!
:
f .g.x// D f 0 .g.x//g 0 .x/ D 3
!
dx
xC1
.x C 1/2
.x C 1/4
d
3x 2
1
2
.3x
/
D
!
:
g.f .x// D g 0 .f .x//f 0 .x/ D ! 3
dx
.x C 1/2
.x 3 C 1/2
S E C T I O N 3.7
The Chain Rule
271
In Exercises 29–42, use the Chain Rule to find the derivative.
29. y D sin.x 2 /
% &
% &
% &
Let y D sin x 2 . Then y 0 D cos x 2 " 2x D 2x cos x 2 .
SOLUTION
30. y D sin2 x
Let y D sin2 x D .sin x/2 . Then y 0 D 2 sin x.cos x/:
SOLUTION
31. y D
p
t2 C 9
SOLUTION
Let y D
p
t 2 C 9 D .t 2 C 9/1=2 . Then
y0 D
32. y D .t 2 C 3t C 1/"5=2
%
&"5=2
SOLUTION Let y D t 2 C 3t C 1
. Then
y0 D !
1 2
t
.t C 9/"1=2 .2t/ D p
:
2
2
t C9
$"7=2
5#2
5 .2t C 3/
t C 3t C 1
.2t C 3/ D ! %
&7=2 :
2
2 t 2 C 3t C 1
33. y D .x 4 ! x 3 ! 1/2=3
%
&2=3
SOLUTION Let y D x 4 ! x 3 ! 1
. Then
y0 D
$"1=3 #
$
2# 4
x ! x3 ! 1
4x 3 ! 3x 2 :
3
p
34. y D . x C 1 ! 1/3=2
#
$3=2
p
1=2
SOLUTION Let y D .x C 1/
!1
. Here, we note that calculating the derivative of the inside function, x C 1 ! 1,
requires the chain rule (in the form of the scaling and shifting rule). After two applications of the chain rule, we have
pp
"
$1=2 ! 1
3#
3
x C1!1
y0 D
.x C 1/1=2 ! 1
"
.x C 1/"1=2 " 1 D
p
:
2
2
4 xC1
!
"
xC1 4
35. y D
x!1
!
"
xC1 4
SOLUTION Let y D
. Then
x!1
y0 D 4
!
xC1
x !1
"3
"
.x ! 1/ " 1 ! .x C 1/ " 1
3
.x ! 1/
2
D!
8 .x C 1/3
.x ! 1/
5
D
8.1 C x/3
:
.1 ! x/5
36. y D cos .12!/
SOLUTION
37. y D sec
SOLUTION
After two applications of the chain rule,
1
x
%
&
Let f .x/ D sec x "1 . Then
38. y D tan.! 2 ! 4!/
SOLUTION
y 0 D 3 cos2 .12!/.! sin.12!//.12/ D !36 cos2 .12!/ sin.12!/:
#
$
#
$ #
$
sec .1=x/ tan .1=x/
f 0 .x/ D sec x "1 tan x "1 " !x "2 D !
:
x2
Let y D tan.! 2 ! 4!/. Then
y 0 D sec2 .! 2 ! 4!/ " .2! ! 4/ D .2! ! 4/ sec2 .! 2 ! 4!/:
39. y D tan.! C cos !/
SOLUTION
Let y D tan .! C cos !/. Then
y 0 D sec2 .! C cos !/ " .1 ! sin !/ D .1 ! sin !/ sec2 .! C cos !/ :
272
DIFFERENTIATION
CHAPTER 3
40. y D e 2x
SOLUTION
2
2
Let y D e 2x . Then
2
41. y D e 2"9t
SOLUTION
2
y 0 D e 2x .4x/ D 4xe 2x :
2
2
Let y D e 2"9t . Then
2
42. y D cos3 .e 4! /
SOLUTION
2
y 0 D e 2"9t .!18t/ D !18te 2"9t :
Let y D cos3 .e 4! /. After two applications of the chain rule, we have
y 0 D 3 cos2 .e 4! /.! sin.e 4! //.4e 4! / D !12e 4! cos2 .e 4! / sin.e 4! /:
In Exercises 43–72, find the derivative using the appropriate rule or combination of rules.
43. y D tan.x 2 C 4x/
SOLUTION
Let y D tan.x 2 C 4x/. By the chain rule,
y 0 D sec2 .x 2 C 4x/ " .2x C 4/ D .2x C 4/ sec2 .x 2 C 4x/:
44. y D sin.x 2 C 4x/
SOLUTION
Let y D sin.x 2 C 4x/. By the chain rule,
dy
D .2x C 4/ cos.x 2 C 4x/:
dx
45. y D x cos.1 ! 3x/
SOLUTION
Let y D x cos .1 ! 3x/. Applying the product rule and then the scaling and shifting rule,
y 0 D x .! sin .1 ! 3x// " .!3/ C cos .1 ! 3x/ " 1 D 3x sin .1 ! 3x/ C cos .1 ! 3x/ :
46. y D sin.x 2 / cos.x 2 /
SOLUTION
We start by using a trig identity to rewrite
y D sin.x 2 / cos.x 2 / D
1
sin.2x 2 /:
2
Then, by the chain rule,
y0 D
1
cos.2x 2 / " 4x D 2x cos.2x 2 /:
2
47. y D .4t C 9/1=2
SOLUTION
Let y D .4t C 9/1=2 . By the shifting and scaling rule,
! "
dy
1
D4
.4t C 9/"1=2 D 2.4t C 9/"1=2 :
dt
2
48. y D .z C 1/4 .2z ! 1/3
SOLUTION
Let y D .z C 1/4 .2z ! 1/3 . Applying the product rule and the general power rule,
dy
D .z C 1/4 .3.2z ! 1/2 /.2/ C .2z ! 1/3 .4.z C 1/3 /.1/ D .z C 1/3 .2z ! 1/2 .6.z C 1/ C 4.2z ! 1//
dz
D .z C 1/3 .2z ! 1/2 .14z C 2/:
49. y D .x 3 C cos x/"4
SOLUTION
Let y D .x 3 C cos x/"4 . By the general power rule,
y 0 D !4.x 3 C cos x/"5 .3x 2 ! sin x/ D 4.sin x ! 3x 2 /.x 3 C cos x/"5 :
50. y D sin.cos.sin x//
S E C T I O N 3.7
51. y D
273
Let y D sin .cos .sin x//. Applying the chain rule twice,
SOLUTION
p
The Chain Rule
y 0 D cos .cos .sin x// " .! sin .sin x// " cos x D ! cos x sin .sin x/ cos .cos .sin x// :
sin x cos x
SOLUTION
We start by using a trig identity to rewrite
yD
p
sin x cos x D
r
1
1
sin 2x D p .sin 2x/1=2 :
2
2
Then, after two applications of the chain rule,
1 1
cos 2x
y 0 D p " .sin 2x/"1=2 " cos 2x " 2 D p
:
2 sin 2x
2 2
52. y D .9 ! .5 ! 2x 4 /7 /3
SOLUTION
Let y D .9 ! .5 ! 2x 4 /7 /3 . Applying the chain rule twice, we find
y 0 D 3.9 ! .5 ! 2x 4 /7 /2 .!7.5 ! 2x 4 /6 /.!8x 3 / D 168x 3 .5 ! 2x 4 /6 .9 ! .5 ! 2x 4 /7 /2 :
53. y D .cos 6x C sin x 2 /1=2
Let y D .cos 6x C sin.x 2 //1=2 . Applying the general power rule followed by both the scaling and shifting rule and
the chain rule,
SOLUTION
y0 D
54. y D
.x C 1/1=2
xC2
SOLUTION
Let y D
&"1=2 %
&
1%
x cos.x 2 / ! 3 sin 6x
cos 6x C sin.x 2 /
! sin 6x " 6 C cos.x 2 / " 2x D p
:
2
cos 6x C sin.x 2 /
.xC1/1=2
xC2 .
Applying the quotient rule and the shifting and scaling rule, we get
.x C 2/ 12 .x C 1/"1=2 ! .x C 1/1=2
dy
1
.x C 2/ ! 2.x C 1/
1
x
D
D p
D! p
:
dx
.x C 2/2
.x C 2/2
2 xC1
2 x C 1 .x C 2/2
55. y D tan3 x C tan.x 3 /
Let y D tan3 x C tan.x 3 / D .tan x/3 C tan.x 3 /. Applying the general power rule to the first term and the chain rule
to the second term,
%
&
y 0 D 3.tan x/2 sec2 x C sec2 .x 3 / " 3x 2 D 3 x 2 sec2 .x 3 / C sec2 x tan2 x :
p
56. y D 4 ! 3 cos x
SOLUTION
SOLUTION
57. y D
r
SOLUTION
Let y D .4 ! 3 cos x/1=2 . By the general power rule,
y0 D
zC1
z!1
Let y D
!
zC1
z!1
"1=2
1
3 sin x
.4 ! 3 cos x/"1=2 " 3 sin x D p
:
2
2 4 ! 3 cos x
. Applying the general power rule followed by the quotient rule,
dy
1
D
dz
2
!
zC1
z!1
""1=2
"
.z ! 1/ " 1 ! .z C 1/ " 1
.z ! 1/2
D p
!1
z C 1 .z ! 1/3=2
:
58. y D .cos3 x C 3 cos x C 7/9
%
&9
SOLUTION Let y D cos3 x C 3 cos x C 7 . Applying the general power rule followed by the sum rule, with the first term
requiring the general power rule,
$8 #
$
#
dy
3 cos2 x " .! sin x/ ! 3 sin x
D 9 cos3 x C 3 cos x C 7
dx
#
$8 #
$
D !27 sin x cos3 x C 3 cos x C 7
1 C cos2 x :
274
DIFFERENTIATION
CHAPTER 3
59. y D
cos.1 C x/
1 C cos x
SOLUTION
Let
yD
cos.1 C x/
:
1 C cos x
Then, applying the quotient rule and the shifting and scaling rule,
dy
!.1 C cos x/ sin.1 C x/ C cos.1 C x/ sin x
cos.1 C x/ sin x ! cos x sin.1 C x/ ! sin.1 C x/
D
D
dx
.1 C cos x/2
.1 C cos x/2
D
sin.!1/ ! sin.1 C x/
:
.1 C cos x/2
The last line follows from the identity
sin.A ! B/ D sin A cos B ! cos A sin B
with A D x and B D 1 C x.
p
60. y D sec. t 2 ! 9/
#p
$
SOLUTION Let y D sec
t 2 ! 9 . Applying the chain rule followed by the general power rule,
#p
$
#p
$ 1#
$"1=2
t sec
dy
D sec
t 2 ! 9 tan
t2 ! 9 "
t2 ! 9
" 2t D
dt
2
61. y D cot7 .x 5 /
SOLUTION
62. y D
% &
Let y D cot7 x 5 . Applying the general power rule followed by the chain rule,
cos.1=x/
1 C x2
SOLUTION
#p
$
#p
$
t 2 ! 9 tan
t2 ! 9
p
:
t2 ! 9
Let y D
# $ #
# $$
# $
# $
dy
D 7 cot6 x 5 " ! csc2 x 5 " 5x 4 D !35x 4 cot6 x 5 csc2 x 5 :
dx
cos.1=x/
1Cx 2
D
cos.x !1 /
.
1Cx 2
Then, applying the quotient rule and the chain rule, we get:
dy
.1 C x 2 /.x "2 sin.x "1 // ! cos.x "1 /.2x/
sin.x "1 / ! 2x cos.x "1 / C x "2 sin.x "1 /
D
D
:
2
2
dx
.1 C x /
.1 C x 2 /2
#
$9
63. y D 1 C cot5 .x 4 C 1/
SOLUTION
succession,
%
%
&&9
Let y D 1 C cot5 x 4 C 1 . Applying the general power rule, the chain rule, and the general power rule in
64. y D 4e "x C 7e "2x
SOLUTION
#
#
$$8
#
$ #
#
$$
dy
D 9 1 C cot5 x 4 C 1
" 5 cot4 x 4 C 1 " ! csc2 x 4 C 1 " 4x 3
dx
#
$
#
$#
#
$$8
D !180x 3 cot4 x 4 C 1 csc2 x 4 C 1 1 C cot5 x 4 C 1
:
Let y D 4e "x C 7e "2x . Using the chain rule twice, once for each exponential function, we obtain
dy
D !4e "x ! 14e "2x :
dx
65. y D .2e 3x C 3e "2x /4
Let y D .2e 3x C 3e "2x /4 . Applying the general power rule followed by two applications of the chain rule, one for
each exponential function, we find
SOLUTION
dy
D 4.2e 3x C 3e "2x /3 .6e 3x ! 6e "2x / D 24.2e 3x C 3e "2x /3 .e 3x ! e "2x /:
dx
66. y D cos.te "2t /
SOLUTION
Let y D cos.te "2t /. Applying the chain rule and the product rule, we have
#
$
dy
D ! sin.te "2t / !2te "2t C e "2t D e "2t .2t ! 1/ sin.te "2t /:
dt
S E C T I O N 3.7
67. y D e .x
SOLUTION
The Chain Rule
2 C2xC3/2
Let y D e .x
2 C2xC3/2
. By the chain rule and the general power rule, we obtain
dy
2
2
2
2
D e .x C2xC3/ " 2.x 2 C 2x C 3/.2x C 2/ D 4.x C 1/.x 2 C 2x C 3/e .x C2xC3/ :
dx
68. y D e e
x
x
Let y D e e . Applying the chain rule, we have
SOLUTION
r
dy
x
D ee ex :
dx
q
p
1C x
!
#
$1=2 "1=2
SOLUTION Let y D 1 C 1 C x 1=2
. Applying the general power rule twice,
69. y D
1C
!
#
$1=2 ""1=2 1 #
$"1=2 1
dy
1
D
1 C 1 C x 1=2
"
1 C x 1=2
" x "1=2 D
dx
2
2
2
q
p p
p
8 x 1C x
p
xC1C1
#
$1=2
1=2
SOLUTION Let y D 1 C .x C 1/
. Applying the general power rule twice,
70. y D
1
:
q
p
p
1C 1C x
$"1=2 1
dy
1#
1
D
1 C .x C 1/1=2
" .x C 1/"1=2 " 1 D p
p
:
p
dx
2
2
4 xC1 1C xC1
71. y D .kx C b/"1=3 ;
SOLUTION
k and b any constants
Let y D .kx C b/"1=3 , where b and k are constants. By the scaling and shifting rule,
1
k
y 0 D ! .kx C b/"4=3 " k D ! .kx C b/"4=3 :
3
3
72. y D p
SOLUTION
1
kt 4 C b
;
k; b constants, not both zero
%
&"1=2
Let y D kt 4 C b
, where b and k are constants. By the chain rule,
y0 D !
$"3=2
1# 4
2kt 3
kt C b
" 4kt 3 D ! %
&3=2 :
2
kt 4 C b
In Exercises 73–76, compute the higher derivative.
d2
sin.x 2 /
dx 2
% &
% &
SOLUTION Let f .x/ D sin x 2 . Then, by the chain rule, f 0 .x/ D 2x cos x 2 and, by the product rule and the chain rule,
#
# $
$
# $
# $
# $
f 00 .x/ D 2x ! sin x 2 " 2x C 2 cos x 2 D 2 cos x 2 ! 4x 2 sin x 2 :
73.
d2 2
.x C 9/5
dx 2
SOLUTION Let f .x/ D .x 2 C 9/5 . Then, by the general power rule,
74.
f 0 .x/ D 5.x 2 C 9/4 " 2x D 10x.x 2 C 9/4
and, by the product rule and the general power rule,
f 00 .x/ D 10x " 4.x 2 C 9/3 " 2x C 10.x 2 C 9/4 D 80x 2 .x 2 C 9/3 C 10.x 2 C 9/4 :
d3
.9 ! x/8
dx 3
8
SOLUTION Let f .x/ D .9 ! x/ . Then, by repeated use of the scaling and shifting rule,
75.
f 0 .x/ D 8.9 ! x/7 " .!1/ D !8.9 ! x/7
f 00 .x/ D !56.9 ! x/6 " .!1/ D 56.9 ! x/6 ;
f 000 .x/ D 336.9 ! x/5 " .!1/ D !336.9 ! x/5 :
275
276
76.
CHAPTER 3
DIFFERENTIATION
d3
sin.2x/
dx 3
SOLUTION
Let f .x/ D sin .2x/. Then, by repeated use of the scaling and shifting rule,
f 0 .x/ D 2 cos.2x/
f 00 .x/ D !4 sin.2x/
f 000 .x/ D !8 cos.2x/:
p
77. The average molecular velocity v of a gas in a certain container is given by v D 29 T
ˇ m/s, where T is the temperature in
dv ˇˇ
kelvins. The temperature is related to the pressure (in atmospheres) by T D 200P . Find
.
ˇ
dP ˇ
SOLUTION
P D1:5
First note that when P D 1:5 atmospheres, T D 200.1:5/ D 300K. Thus,
p
ˇ
d T ˇˇ
29
290 3
m
D
p
"
200
D
:
ˇ
dP
3
s
"
atmospheres
2
300
T D300
P D1:5
p
Alternately, substituting T D 200P into the equation for v gives v D 290 2P . Therefore,
p
dv
290 2
290
D p
D p ;
dP
2 P
2P
ˇ
dv ˇˇ
dv
D
dP ˇP D1:5
dT
ˇ
ˇ
ˇ
ˇ
"
so
p
ˇ
dv ˇˇ
290
290 3
m
D
p
D
:
dP ˇP D1:5
3 s " atmospheres
3
78. The power P in a circuit is P D Ri 2 , where R is the resistance and i is the current. Find dP =dt at t D
and i varies according to i D sin.4# t/ (time in seconds).
ˇ
ˇ
p
d # 2 $ˇˇ
di ˇ
SOLUTION
Ri ˇ
D 2Ri ˇˇ
D 2.1000/4# sin.4# t/ cos.4# t/jt D1=3 D 2000# 3.
dt
dt
t D1=3
1
3
if R D 1000 %
t D2
79. An expanding sphere has radius r D 0:4t cm at time t (in seconds). Let V be the sphere’s volume. Find dV =dt when
(a) r D 3 and (b) t D 3.
SOLUTION
Let r D 0:4t, where t is in seconds (s) and r is in centimeters (cm). With V D 43 # r 3 , we have
dV
D 4# r 2 :
dr
Thus
dV
dV dr
D
D 4# r 2 " .0:4/ D 1:6# r 2 :
dt
dr dt
dV
D 1:6#.3/2 # 45:24 cm/s.
dt
dV
(b) When t D 3; we have r D 1:2. Hence
D 1:6#.1:2/2 # 7:24 cm/s.
dt
80. A 2005 study by the Fisheries Research Services in Aberdeen, Scotland, suggests that the average length of the species Clupea
harengus (Atlantic herring) as a function of age t (in years) can be modeled by L.t/ D 32.1 ! e "0:37t / cm for 0 $ t $ 13. See
Figure 1.
(a) How fast is the average length changing at age t D 6 years?
(b) At what age is the average length changing at a rate of 5 cm/yr?
(a) When r D 3,
L (cm)
32
30
20
10
2
4
6
8
10
t (year)
FIGURE 1 Average length of the species Clupea harengus
12
S E C T I O N 3.7
SOLUTION
The Chain Rule
277
Let L.t/ D 32.1 ! e "0:37t /. Then
L0 .t/ D 32.0:37/e "0:37t D 11:84e "0:37t :
(a) At age t D 6,
L0 .t/ D 11:84e "0:37.6/ D 11:84e "2:22 # 1:29 cm=yr:
(b) The length will be changing at a rate of 5 cm/yr when
11:84e "0:37t D 5:
Solving for t yields
1
5
ln
# 2:33 years:
0:37 11:84
81. A 1999 study by Starkey and Scarnecchia developed the following model for the average weight (in kilograms) at age t (in
years) of channel catfish in the Lower Yellowstone River (Figure 2):
t D!
W .t/ D .3:46293 ! 3:32173e "0:03456t /3:4026
Find the rate at which average weight is changing at age t D 10.
W (kg)
8
7
6
5
4
3
2
1
Lower Yellowstone River
5
10
15
20
t (year)
FIGURE 2 Average weight of channel catfish at age t
SOLUTION
Let W .t/ D .3:46293 ! 3:32173e "0:03456t /3:4026 . Then
W 0 .t/ D 3:4026.3:46293 ! 3:32173e "0:03456t /2:4026 .3:32173/.0:03456/e "0:03456t
D 0:3906.3:46293 ! 3:32173e "0:03456t /2:4026 e "0:03456t :
At age t D 10,
W 0 .10/ D 0:3906.1:1118/2:4026 .0:7078/ # 0:3566 kg=yr:
82. The functions in Exercises 80 and 81 are examples of the von Bertalanffy growth function
%
&1=m
M.t/ D a C .b ! a/e kmt
.m ¤ 0/
%
&1=m
M.t/ D a C .b ! a/e kmt
.m ¤ 0/:
introduced in the 1930s by Austrian-born biologist Karl Ludwig von Bertalanffy. Calculate M 0 .0/ in terms of the constants a, b, k
and m.
SOLUTION
Let
Then
M 0 .t/ D
1
.a C .b ! a/e kmt /1=m"1 km.b ! a/e kmt D k.b ! a/e kmt .a C .b ! a/e kmt /1=m"1 ;
m
and
M 0 .0/ D k.b ! a/e 0 .a C .b ! a/e 0 /1=m"1 D k.b ! a/b 1=m"1 :
83. With notation as in Example 7, calculate
ˇ
ˇ
d
(a)
sin ! ˇˇ
d!
! D60ı
(b)
ˇ
ˇ
d
.! C tan !/ ˇˇ
d!
! D45ı
278
CHAPTER 3
DIFFERENTIATION
SOLUTION
(a)
(b)
ˇ
# # $ˇ
# # $
# #
$
d
d
# 1
#
ˇ
ˇ
sin ! ˇ
D
sin
! ˇ
D
cos
.60/ D
D
:
ı
ı
! D60
! D60
d!
d!
180
180
180
180 2
360
ˇ
# # $$ ˇ
# $
d
d #
#
#
ˇ
ˇ
2 #
.! C tan !/ ˇ
D
!
C
tan
!
D
1
C
sec
D 1C :
ˇ
! D45ı d!
! D45ı
d!
180
180
4
90
84. Assume that
f .0/ D 2; f 0 .0/ D 3; h.0/ D !1; h0 .0/ D 7
Calculate the derivatives of the following functions at x D 0:
(a) .f .x//3
(b) f .7x/
(c) f .4x/h.5x/
SOLUTION
(a) Let g.x/ D .f .x//3 . Then
g 0 .0/ D 3.f .0//2 .f 0 .0// D 12.3/ D 36:
(b) Let g.x/ D f .7x/. Then
g 0 .0/ D 7f 0 .7.0// D 21:
(c) Let F .x/ D f .4x/h.5x/. Then F 0 .x/ D 4f 0 .4x/h.5x/ C 5f .4x/h0 .5x/ and
F 0 .0/ D 4.3/.!1/ C 5.2/.7/ D 58:
85. Compute the derivative of h.sin x/ at x D
SOLUTION
"
6,
assuming that h0 .0:5/ D 10.
Let u D sin x and suppose that h0 .0:5/ D 10. Then
d
dh du
dh
.h.u// D
D
cos x:
dx
du dx
du
When x D
"
6;
we have u D :5. Accordingly, the derivative of h.sin x/ at x D
"
6
is 10 cos
86. Let F .x/ D f .g.x//, where the graphs of f and g are shown in Figure 3. Estimate
4
%" &
p
6 D5
0
g .2/ and
3:
f 0 .g.2// and compute F 0 .2/.
y
3
g(x)
f(x)
2
1
x
1
2
3
4
5
FIGURE 3
After sketching an approximate tangent line to g at x D 2 (see the figure below), we estimate g 0 .2/ D !1. It appears
from the graph that g.2/ D 3 and f 0 .3/ D 54 (since between x D 2 and x D 4 the graph of f appears to be linear with slope 54 ).
Thus,
SOLUTION
F 0 .2/ D f 0 .g.2//g 0 .2/ D
4
5
.!1/ D !1:25:
4
y
3
2
1
x
1
2
3
4
5
S E C T I O N 3.7
The Chain Rule
279
In Exercises 87–90, use the table of values to calculate the derivative of the function at the given point.
x
1
4
6
f .x/
f 0 .x/
g.x/
g 0 .x/
4
5
4
5
0
7
1
6
4
6
3
1
2
87. f .g.x//, x D 6
SOLUTION
ˇ
ˇ
d
f .g.x//ˇˇ
D f 0 .g.6//g 0 .6/ D f 0 .6/g 0 .6/ D 4 % 3 D 12.
dx
xD6
88. e f .x/ , x D 4
ˇ
d f .x/ ˇˇ
SOLUTION
e
D e f .4/ f 0 .4/ D e 0 .7/ D 7.
ˇ
dx
xD4
p
89. g. x/, x D 16
ˇ
! "
! "! "! "
p
d p ˇˇ
1
1
1
1
1
SOLUTION
g. x/ˇ
D g 0 .4/
.1= 16/ D
D
.
dx
2
2
2
4
16
xD16
90. f .2x C g.x//, x D 1
SOLUTION
ˇ
ˇ
d
f .2x C g.x//ˇˇ
D f 0 .2.1/ C g.1//.2 C g 0 .1// D f 0 .2 C 4/.7/ D 4.7/ D 28.
dx
xD1
91. The price (in dollars) of a computer component is P D 2C ! 18C "1 , where C is the manufacturer’s cost to produce it.
Assume that cost at time t (in years) is C D 9 C 3t "1 . Determine the rate of change of price with respect to time at t D 3.
SOLUTION
dC
D !3t "2 . C.3/ D 10 and C 0 .3/ D ! 13 , so we compute:
dt
ˇ
! "
dP ˇˇ
18
2
18
1
dollars
0
0
D
2C
.3/
C
C
.3/
D
!
C
!
D !0:727
:
dt ˇt D3
3
100
3
year
.C.3//2
92.
Plot the “astroid” y D .4 ! x 2=3 /3=2 for 0 $ x $ 8. Show that the part of every tangent line in the first quadrant has
a constant length 8.
SOLUTION
!
Here is a graph of the astroid.
y
10
10
−10
x
−10
!
Let f .x/ D .4 ! x 2=3 /3=2 . Then
f 0 .x/ D
p
!
"
3
2
4 ! x 2=3
.4 ! x 2=3 /1=2 ! x "1=3 D !
;
2
3
x 1=3
and the tangent line to f at x D a is
p
#
$3=2
4 ! a2=3
2=3
yD!
.x
!
a/
C
4
!
a
:
a1=3
#
$
% p
&
The y-intercept of this line is the point P D 0; 4 4 ! a2=3 , its x-intercept is the point Q D 4a1=3 ; 0 , and the distance
between P and Q is 8.
93. According to the U.S. standard atmospheric model, developed by the National Oceanic and Atmospheric Administration for
use in aircraft and rocket design, atmospheric temperature T (in degrees Celsius), pressure P (kPa D 1;000 pascals), and altitude
h (in meters) are related by these formulas (valid in the troposphere h $ 11;000):
!
"
T C 273:1 5:256
T D 15:04 ! 0:000649h;
P D 101:29 C
288:08
280
DIFFERENTIATION
CHAPTER 3
Use the Chain Rule to calculate dP =dh. Then estimate the change in P (in pascals, Pa) per additional meter of altitude when
h D 3;000.
SOLUTION
!
"
!
"
dP
T C 273:1 4:256
1
D 5:256
D 6:21519 % 10"13 .273:1 C T /4:256
dT
288:08
288:08
and
dT
dh
D !0:000649ı C=m.
dP
dh
D
dP d T
d T dh
, so
#
$
dP
D 6:21519 % 10"13 .273:1 C T /4:256 .!0:000649/ D !4:03366 % 10"16 .288:14 ! 0:000649 h/4:256 :
dh
When h D 3000,
dP
D !4:03366 % 10"16 .286:193/4:256 D !1:15 % 10"5 kPa=mI
dh
therefore, for each additional meter of altitude,
"P # !1:15 % 10"5 kPa D !1:15 % 10"2 Pa:
94. Climate scientists use the Stefan-Boltzmann Law R D &T 4 to estimate the change in the earth’s average temperature T
(in kelvins) caused by a change in the radiation R (in joules per square meter per second) that the earth receives from the sun.
Here & D 5:67 % 10"8 Js"1 m"2 K"4 . Calculate dR=dt , assuming that T D 283 and ddtT D 0:05 K/yr. What are the units of the
derivative?
SOLUTION
By the Chain Rule,
dR
dR d T
dT
D
"
D 4&T 3
:
dt
d T dt
dt
Assuming T D 283 K and
dT
dt
D 0:05 K/yr, it follows that
dR
D 4&.2833 /.0:05/ # 0:257 Js"1 m"2 /yr
dt
95. In the setting of Exercise 94, calculate the yearly rate of change of T if T D 283 K and R increases at a rate of 0:5 Js"1 m"2
per year.
SOLUTION
By the Chain Rule,
dR
dR d T
dT
D
"
D 4&T 3
:
dt
d T dt
dt
Assuming T D 283 K and
dR
dt
D 0:5 Js"1 m"2 per year, it follows that author:
0:5 D 4&.283/3
dT
dT
0:5
)
D
# 0:0973 kelvins/yr
dt
dt
4&.283/3
Use a computer algebra system to compute f .k/ .x/ for k D 1; 2; 3 for the following functions:
p
(a) f .x/ D cot.x 2 /
(b) f .x/ D x 3 C 1
96.
SOLUTION
(a) Let f .x/ D cot.x 2 /. Using a computer algebra system,
f 0 .x/ D !2x csc2 .x 2 /I
f 00 .x/ D 2 csc2 .x 2 /.4x 2 cot.x 2 / ! 1/I and
#
$
f 000 .x/ D !8x csc2 .x 2 / 6x 2 cot2 .x 2 / ! 3 cot.x 2 / C 2x 2 :
(b) Let f .x/ D
p
x 3 C 1. Using a computer algebra system,
3x 2
f 0 .x/ D p
I
2 x3 C 1
f 00 .x/ D
3x.x 3 C 4/
I and
4.x 3 C 1/3=2
f 000 .x/ D !
3.x 6 C 20x 3 ! 8/
:
8.x 3 C 1/5=2
S E C T I O N 3.7
The Chain Rule
281
97. Use the Chain Rule to express the second derivative of f ı g in terms of the first and second derivatives of f and g.
SOLUTION
Let h.x/ D f .g.x//. Then
h0 .x/ D f 0 .g.x//g 0.x/
and
%
&2
h00 .x/ D f 0 .g.x//g 00 .x/ C g 0 .x/f 00 .g.x//g 0.x/ D f 0 .g.x//g 00.x/ C f 00 .g.x// g 0 .x/ :
98. Compute the second derivative of sin.g.x// at x D 2, assuming that g.2/ D
SOLUTION
"
4,
Let f .x/ D sin.g.x//. Then f 0 .x/ D cos.g.x//g 0 .x/ and
g 0 .2/ D 5, and g 00 .2/ D 3.
f 00 .x/ D cos.g.x//g 00 .x/ C g 0 .x/.! sin.g.x///g 0.x/ D cos.g.x//g 00 .x/ ! .g 0 .x//2 sin.g.x//:
Therefore,
%
&2
% &
% &
f 00 .2/ D g 00 .2/ cos .g.2// ! g 0 .2/ sin .g.2// D 3 cos "4 ! .5/2 sin "4 D !22 "
p
2
2
p
D !11 2
Further Insights and Challenges
99. Show that if f , g, and h are differentiable, then
Œf .g.h.x///$0 D f 0 .g.h.x///g 0 .h.x//h0 .x/
SOLUTION
Let f , g, and h be differentiable. Let u D h.x/, v D g.u/, and w D f .v/. Then
dw
df dv
df dg du
D
D
D f 0 .g.h.x//g 0 .h.x//h0 .x/
dx
dv dx
dv du dx
100.
Show that differentiation reverses parity: If f is even, then f 0 is odd, and if f is odd, then f 0 is even. Hint: Differentiate f .!x/.
SOLUTION A function is even if f .!x/ D f .x/ and odd if f .!x/ D !f .x/. By the chain rule,
suppose that f is even. Then f .!x/ D f .x/ and
d
f .!x/
dx
D !f 0 .!x/. Now
d
d
f .!x/ D
f .x/ D f 0 .x/:
dx
dx
Hence, when f is even, !f 0 .!x/ D f 0 .x/ or f 0 .!x/ D !f 0 .x/ and f 0 is odd. On the other hand, suppose f is odd. Then
f .!x/ D !f .x/ and
d
d
f .!x/ D ! f .x/ D !f 0 .x/:
dx
dx
Hence, when f is odd, !f 0 .!x/ D !f 0 .x/ or f 0 .!x/ D f 0 .x/ and f 0 is even.
101. (a)
Sketch a graph of any even function f .x/ and explain graphically why f 0 .x/ is odd.
(b) Suppose that f 0 .x/ is even. Is f .x/ necessarily odd? Hint: Check whether this is true for linear functions.
SOLUTION
(a) The graph of an even function is symmetric with respect to the y-axis. Accordingly, its image in the left half-plane is a mirror
reflection of that in the right half-plane through the y-axis. If at x D a & 0, the slope of f exists and is equal to m, then by
reflection its slope at x D !a $ 0 is !m. That is, f 0 .!a/ D !f 0 .a/. Note: This means that if f 0 .0/ exists, then it equals 0.
y
4
3
2
1
−2
−1
1
2
x
(b) Suppose that f 0 is even. Then f is not necessarily odd. Let f .x/ D 4x C 7. Then f 0 .x/ D 4, an even function. But f is not
odd. For example, f .2/ D 15, f .!2/ D !1, but f .!2/ ¤ !f .2/.
102. Power Rule for Fractional Exponents Let f .u/ D uq and g.x/ D x p=q . Assume that g.x/ is differentiable.
(a) Show that f .g.x// D x p (recall the laws of exponents).
(b) Apply the Chain Rule and the Power Rule for whole-number exponents to show that f 0 .g.x// g 0 .x/ D px p"1 .
(c) Then derive the Power Rule for x p=q .
282
CHAPTER 3
DIFFERENTIATION
SOLUTION
(a) Let f .u/ D uq and g.x/ D x p=q , where q is a positive integer and p is an integer. Then
#
$ #
$q
f .g.x// D f x p=q D x p=q D x p :
(b) Differentiating both sides of the final expression in part (a), applying the chain rule on the left and the power rule for whole
number exponents on the right, it follows that
f 0 .g.x//g 0 .x/ D px p"1 :
(c) Thus
g 0 .x/ D
px p"1
px p"1
px p"1
p
D
D
D x p=q"1 :
%
&
q"1
p"p=q
f 0 .g.x//
q
p=q
qx
q x
103. Prove that for all whole numbers n & 1,
%
Hint: Use the identity cos x D sin x C
SOLUTION
"
2
&
.
#
dn
n# $
sin x D sin x C
n
dx
2
We will proceed by induction on n. For n D 1, we find
#
d
#$
sin x D cos x D sin x C
;
dx
2
as required. Now, suppose that for some positive integer k,
#
dk
k# $
sin x D sin x C
:
2
dx k
Then
#
d kC1
d
k# $
sin x D
sin x C
kC1
dx
2
dx
!
"
!
"
k#
.k C 1/#
D cos x C
D sin x C
:
2
2
104. A Discontinuous Derivative
SOLUTION
Use the limit definition to show that g 0 .0/ exists but g 0 .0/ ¤ lim g 0 .x/, where
Using the limit definition,
x!0
8̂
1
<x 2 sin
x
g.x/ D
:̂0
x¤0
xD0
# $
! "
h2 sin h1 ! 0
g.0
C
h/
!
g.0/
1
g 0 .0/ D lim
D lim
D lim h sin
D 0;
h
h
h
h!0
h!0
h!0
where we have used the squeeze theorem in the last step. Now, for x ¤ 0,
!
"
! "
! "
! "
! "
1
1
1
1
1
0
2
g .x/ D x ! 2 cos
C 2x sin
D 2x sin
! cos
:
x
x
x
x
x
Although the first term in g 0 has a limit of 0 as x ! 0 (by the squeeze theorem), the limit as x ! 0 of the second term does not
exist. Hence, limx!0 g 0 .x/ does not exist, so g 0 .0/ ¤ limx!0 g 0 .x/.
105. Chain Rule
a new function
This exercise proves the Chain Rule without the special assumption made in the text. For any number b, define
F .u/ D
f .u/ ! f .b/
u!b
for all u ¤ b
(a) Show that if we define F .b/ D f 0 .b/, then F .u/ is continuous at u D b.
(b) Take b D g.a/. Show that if x ¤ a, then for all u,
Note that both sides are zero if u D g.a/.
u ! g.a/
f .u/ ! f .g.a//
D F .u/
x!a
x!a
2
S E C T I O N 3.8
Derivatives of Inverse Functions
283
(c) Substitute u D g.x/ in Eq. (2) to obtain
f .g.x// ! f .g.a//
g.x/ ! g.a/
D F .g.x//
x!a
x !a
Derive the Chain Rule by computing the limit of both sides as x ! a.
SOLUTION
For any differentiable function f and any number b, define
F .u/ D
f .u/ ! f .b/
u!b
for all u ¤ b.
(a) Define F .b/ D f 0 .b/. Then
lim F .u/ D lim
u!b
u!b
f .u/ ! f .b/
D f 0 .b/ D F .b/;
u!b
i.e., lim F .u/ D F .b/. Therefore, F is continuous at u D b.
u!b
(b) Let g be a differentiable function and take b D g.a/. Let x be a number distinct from a. If we substitute u D g.a/ into Eq.
(2), both sides evaluate to 0, so equality is satisfied. On the other hand, if u ¤ g.a/, then
f .u/ ! f .g.a//
f .u/ ! f .g.a// u ! g.a/
f .u/ ! f .b/ u ! g.a/
u ! g.a/
D
D
D F .u/
:
x!a
u ! g.a/
x!a
u!b
x!a
x!a
Hence for all u, we have
f .u/ ! f .g.a//
u ! g.a/
D F .u/
:
x!a
x!a
(c) Substituting u D g.x/ in Eq. (2), we have
f .g.x// ! f .g.a//
g.x/ ! g.a/
D F .g.x//
:
x!a
x!a
Letting x ! a gives
!
"
f .g.x// ! f .g.a//
g.x/ ! g.a/
lim
D lim F .g.x//
D F .g.a//g 0 .a/ D F .b/g 0 .a/ D f 0 .b/g 0 .a/
x!a
x!a
x !a
x!a
D f 0 .g.a//g 0 .a/
Therefore .f ı g/0 .a/ D f 0 .g.a//g 0 .a/, which is the Chain Rule.
3.8 Derivatives of Inverse Functions
Preliminary Questions
1. What is the slope of the line obtained by reflecting the line y D
SOLUTION
x
2
through the line y D x?
The line obtained by reflecting the line y D x=2 through the line y D x has slope 2.
2. Suppose that P D .2; 4/ lies on the graph of f .x/ and that the slope of the tangent line through P is m D 3. Assuming that
f "1 .x/ exists, what is the slope of the tangent line to the graph of f "1 .x/ at the point Q D .4; 2/?
SOLUTION
The tangent line to the graph of f "1 .x/ at the point Q D .4; 2/ has slope 13 .
3. Which inverse trigonometric function g.x/ has the derivative g 0 .x/ D
SOLUTION
g.x/ D tan"1 x has the derivative g 0 .x/ D
1
.
x2 C 1
1
?
x2 C 1
4. What does the following identity tell us about the derivatives of sin"1 x and cos"1 x?
sin"1 x C cos"1 x D
SOLUTION
Angles whose sine and cosine are x are complementary.
#
2