Section 3.4 The Chain Rule

(1)

Section 3.4 The Chain Rule

46. Find the derivative of the function. y = q

x +p x +√

x.

Solution:

SECTION 3.4 THE CHAIN RULE ¤ 215

35. Using Formula 5 and the Chain Rule, () = 4^ ⇒

⁰() = 4^(ln 4) 









 = ⁻¹



= 4^(ln 4)

−⁻²

= − (ln 4)4^

² .

36.  () =

⁴+ 1

²+ 1

5

⇒

⁰() = 5

⁴+ 1

²+ 1

4

(²+ 1)(4³) − (⁴+ 1)(2)

(²+ 1)² =5(⁴+ 1)⁴2[2²(²+ 1) − (⁴+ 1)]

(²+ 1)⁴(²+ 1)²

= 10(⁴+ 1)⁴(⁴+ 2²− 1) (²+ 1)⁶

37.  () = sin  cos(1 − ²) ⇒

⁰() = sin 

− sin(1 − ²)(−2)

+ cos(1 − ²) · cos  = 2 sin  sin(1 − ²) + cos  cos(1 − ²) 38. () = ^−cos(²) ⇒ ⁰() = ^−[− sin(²)] · 2 + cos(²) · ^−(−1) = −^−[2 sin(²) + cos(²)]

39.  () = tan√

1 + ²⇒ ⁰() = sec²√

1 + ²· 1 2√

1 + ² · 2 =  sec²√ 1 + ²

√1 + ²

40. () = (1 + cos²)³⇒ ⁰() = 3(1 + cos²)²[2(cos )(− sin )] = −6 cos  sin (1 + cos²)² 41.  = sin²(²+ 1) ⇒ ⁰= 2 sin(²+ 1) · cos(²+ 1) · 2 = 4 sin(²+ 1) cos(²+ 1) 42.  = ^{sin 2}+ sin(^2) ⇒

⁰= ^{sin 2} 

sin 2 + cos(^2) 

^2= ^{sin 2}(cos 2) · 2 + cos(^2) ^2· 2 = 2 cos 2 ^{sin 2}+ 2^2cos(^2) 43. () = sin

 ^ 1 + ^



⇒

⁰() = cos

 ^ 1 + ^



·(1 + ^)^− ^(^) (1 + ^)² = cos

 ^ 1 + ^



·^(1 + ^− ^)

(1 + ^)² = ^ (1 + ^)²cos

 ^ 1 + ^



44.  () = ^1√

²− 1 ⇒

⁰() = ^1· 1 2√

²− 1· 2 +√

²− 1 · ^1·



−1

²

 

1

 = ⁻¹; 



⁻¹

= −⁻²= −1

²



= ^1

 

√²− 1−

√²− 1

²

 or ^1

³− ²+ 1

²√

²− 1



45.  () = tan(sec(cos )) ⇒

⁰() = sec²(sec(cos )) 

 sec(cos ) = sec²(sec(cos ))[sec(cos ) tan(cos )] 

 cos 

= − sec²(sec(cos )) sec(cos ) tan(cos ) sin 

46.  =



 +

 +√

 ⇒ ⁰=¹₂

 +

 +√

−12 1 +¹₂

 +√

−12

1 +¹₂^−12

47.  () = ^sin²^(²⁾⇒ ⁰() = ^sin²^(²⁾· 2 sin(²) · cos(²) · 2 = 4 sin(²) cos(²)^sin²^(²⁾

48. Find the derivative of the function. y = 2³^4x. Solution:

SECTION 3.4 THE CHAIN RULE ¤ 201 41.  () = tan(^) + ^{tan } ⇒ ⁰() = sec²(^) · 

(^) + ^{tan }· 

(tan ) = sec²(^) · ^+ ^{tan }· sec²

42.  = sin(sin(sin )) ⇒ ⁰= cos(sin(sin )) 

(sin(sin )) = cos(sin(sin )) cos(sin ) cos  43. () = (2^+ )^ ⇒

⁰() = (2^+ )^⁻¹· 

(2^+ ) = (2^+ )^⁻¹· 2^(ln ) ·  = 2²(ln )(2^+ )^⁻¹^

44.  = 2³^4 ⇒

⁰= 2³^4(ln 2) 

3⁴^ = 2³^4(ln 2) 3⁴^(ln 3) 

4^= 2³^4(ln 2) 3⁴^(ln 3) 4^(ln 4) = (ln 2)(ln 3)(ln 4)4^3⁴^2³^4

45.  = cos

sin(tan ) = cos(sin(tan ))^12 ⇒

⁰= − sin(sin(tan ))^12· 

(sin(tan ))^12= − sin(sin(tan ))^12·¹2(sin(tan ))^−12· 

(sin(tan ))

=− sin

sin(tan ) 2

sin(tan ) · cos(tan ) · 

tan  = − sin

sin(tan ) · cos(tan ) · sec²() · 

=− cos(tan ) sec²() sin

sin(tan )

46.  =

 + ( + sin²)³4

⇒ ⁰= 4

 + ( + sin²)³3

·

1 + 3( + sin²)²· (1 + 2 sin  cos )

47.  = cos(sin 3) ⇒ ⁰= − sin(sin 3) · (cos 3) · 3 = −3 cos 3 sin(sin 3) ⇒

⁰⁰= −3 [(cos 3) cos(sin 3)(cos 3) · 3 + sin(sin 3)(− sin 3) · 3] = −9 cos²(3) cos(sin 3) + 9(sin 3) sin(sin 3)

48.  = 1

(1 + tan )² = (1 + tan )⁻² ⇒ ⁰= −2(1 + tan )⁻³sec² = −2 sec² (1 + tan )³. Using the Product Rule with ⁰=

−2(1 + tan )⁻³

(sec )², we get

⁰⁰= −2(1 + tan )⁻³· 2(sec )(sec  tan ) + (sec )²· 6(1 + tan )⁻⁴sec²

= 2 sec² (1 + tan )⁻⁴

−2(1 + tan ) tan  + 3 sec² 

2 is the lesser exponent for sec  and −4 for (1 + tan )



= 2 sec² (1 + tan )⁻⁴

−2 tan  − 2 tan² + 3(tan² + 1)

=2 sec²

tan² − 2 tan  + 3 (1 + tan )⁴

49.  =√

1 − sec  ⇒ ⁰=¹₂(1 − sec )^−12(− sec  tan ) = − sec  tan  2√

1 − sec . Using the Product Rule with ⁰=

−¹2sec  tan 

(1 − sec )^−12, we get

62. (a) The curve y = ^√^|x|

2−x² is called a bullet-nose curve. Find an equation of the tangent line to this curve at the point (1, 1).

(b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.

Solution:

202 ¤ CHAPTER 3 DIFFERENTIATION RULES

⁰⁰=

−¹2sec  tan 

−¹2(1 − sec )^−32(− sec  tan )

+ (1 − sec )^−12

−¹2

[sec  sec² + tan  sec  tan ].

Now factor out −¹2sec (1 − sec )^−32. Note that −³2 is the lesser exponent on (1 − sec ). Continuing,

⁰⁰= −¹2sec  (1 − sec )^−32₁

2sec  tan² + (1 − sec )(sec² + tan²)

= −¹2sec  (1 − sec )^−321

2sec  tan² + sec² + tan² − sec³ − sec  tan²

= −¹2sec  (1 − sec )^−32

−¹2sec  (sec² − 1) + sec² + (sec² − 1) − sec³

= −¹2sec  (1 − sec )^−32

−³2sec³ + 2 sec² +¹₂sec  − 1

= sec  (1 − sec )^−323

4sec³ − sec² −¹4sec  +¹₂

=sec  (3 sec³ − 4 sec² − sec  + 2) 4(1 − sec )^32

There are many other correct forms of ⁰⁰, such as ⁰⁰=sec  (3 sec  + 2)√ 1 − sec 

4 . We chose to ﬁnd a factored form with only secants in the ﬁnal form.

50.  = ^^ ⇒ ⁰= ^^· (^)⁰= ^^· ^ ⇒

⁰⁰= ^^· (^)⁰+ ^·

^^0

= ^^· ^+ ^· ^^· ^= ^^· ^(1 + ^) or ^^^+(1 + ^)

51.  = 2^ ⇒ ⁰= 2^ln 2. At (0 1), ⁰= 2⁰ln 2 = ln 2, and an equation of the tangent line is  − 1 = (ln 2)( − 0) or  = (ln 2) + 1.

52.  =√

1 + ³= (1 + ³)^12 ⇒ ⁰=¹₂(1 + ³)^−12· 3² = 3² 2√

1 + ³. At (2 3), ⁰= 3 · 4 2√

9= 2, and an equation of the tangent line is  − 3 = 2( − 2), or  = 2 − 1.

53.  = sin(sin ) ⇒ ⁰= cos(sin ) · cos . At ( 0), ⁰= cos(sin ) · cos  = cos(0) · (−1) = 1(−1) = −1, and an equation of the tangent line is  − 0 = −1( − ), or  = − + .

54.  = ^−² ⇒ ⁰= ^−²(−2) + ^−²(1) = ^−²(−2²+ 1). At (0 0), ⁰= ⁰(1) = 1, and an equation of the tangent line is  − 0 = 1( − 0) or  = .

55. (a)  = 2

1 + ^− ⇒ ⁰=(1 + ^−)(0) − 2(−^−)

(1 + ^−)² = 2^−

(1 + ^−)². At (0 1), ⁰= 2⁰

(1 + ⁰)² = 2(1) (1 + 1)² = 2

2² =1

2. So an equation of the tangent line is  − 1 =¹2( − 0) or  = ¹2 + 1.

(b)

56. (a) For   0, || = , and  = () = 

√2 − ² ⇒

⁰() =

√2 − ²(1) − 1 2

(2 − ²)^−12(−2)

√2 − ²² ·(2 − ²)^12 (2 − ²)^12

= (2 − ²) + ²

(2 − ²)^32 = 2 (2 − ²)^32

So at (1 1), the slope of the tangent line is ⁰(1) = 2and its equation is  − 1 = 2( − 1) or  = 2 − 1.

(b)

69. A table of values for f , g, f⁰ and g⁰ is given.

Section3.4 The Chain Rule 205

66. If f is the function whose graph is shown, let hsxd − f s f sxdd and tsxd − f sx²d. Use the graph of f to estimate the value of each derivative.

(a) h9s2d (b) t9s2d

x y

0 1

y=ƒ 1

67. If tsxd − sf sxd, where the graph of f is shown, evaluate t9s3d.

x y

0 1

1

f

68. Suppose f is differentiable on R and is a real number.

Let Fsxd − f sxd and Gsxd − f f sxdg. Find expressions for (a) F9sxd and (b) G9sxd.

69. Suppose f is differentiable on R. Let Fsxd − f se^xd and Gsxd − e^f^sxd. Find expressions for (a) F9sxd and (b) G9sxd.

70. Let tsxd − e^cx1fsxd and hsxd − e^kxfsxd, where f s0d − 3, f 9s0d − 5, and f 99s0d − 22.

(a) Find t9s0d and t99s0d in terms of c.

(b) In terms of k, find an equation of the tangent line to the graph of h at the point where x − 0.

71. Let rsxd − f stshsxddd, where hs1d − 2, ts2d − 3, h9s1d − 4, t9s2d − 5, and f 9s3d − 6. Find r9s1d.

72. If t is a twice differentiable function and f sxd − xtsx²d, find f 99 in terms of t, t9, and t99.

73. If Fsxd − f s3f s4 f sxddd, where f s0d − 0 and f 9s0d − 2, find F9s0d.

74. If Fsxd − f sx f sx f sxddd, where f s1d − 2, f s2d − 3, f 9s1d − 4, f 9s2d − 5, and f 9s3d − 6, find F9s1d.

75. Show that the function y − e^2xsA cos 3x 1 B sin 3xd satisfies the differential equation y99 2 4y9 1 13y − 0.

76. For what values of r does the function y − e^rx satisfy the differential equation y99 2 4y9 1 y − 0?

77. Find the 50th derivative of y − cos 2x.

78. Find the 1000th derivative of fsxd − xe^2x. 56. (a) The curve y −

|

^x

|

^y^{s2 2 x}² is called a bullet-nose

curve. Find an equation of the tangent line to this curve at the point s1, 1d.

(b) Illustrate part (a) by graphing the curve and the tangent line on the same screen.

57. (a) If fsxd − xs2 2 x², find f 9sxd.

(b) Check to see that your answer to part (a) is reasonable by comparing the graphs of f and f 9.

58. The function fsxd − sinsx 1 sin 2xd, 0 < x < , arises in applications to frequency modulation (FM) synthesis.

(a) Use a graph of f produced by a calculator to make a rough sketch of the graph of f 9.

(b) Calculate f 9sxd and use this expression, with a calculator, to graph f 9. Compare with your sketch in part (a).

59. Find all points on the graph of the function

fsxd − 2 sin x 1 sin²x at which the tangent line is horizontal.

60. At what point on the curve y −s1 1 2x is the tangent line perpendicular to the line 6x 1 2y − 1?

61. If Fsxd − f stsxdd, where f s22d − 8, f 9s22d − 4, f 9s5d − 3, ts5d − 22, and t9s5d − 6, find F9s5d.

62. If hsxd −s4 1 3f sxd, where fs1d − 7 and f 9s1d − 4, find h9s1d.

63. A table of values for f , t, f 9, and t9 is given.

x fsxd tsxd f 9sxd t9sxd

1 3 2 4 6

2 1 8 5 7

3 7 2 7 9

(a) If hsxd − f stsxdd, find h9s1d.

(b) If Hsxd −ts f sxdd, find H9s1d.

64. Let f and t be the functions in Exercise 63.

(a) If Fsxd − f s f sxdd, find F9s2d.

(b) If Gsxd −tstsxdd, find G9s3d.

65. If f and t are the functions whose graphs are shown, let usxd − f stsxdd, ^vsxd −ts f sxdd, and ^wsxd −tstsxdd. Find each derivative, if it exists. If it does not exist, explain why.

(a) u9s1d (b) v9s1d (c) w9s1d

x y

0

f

1 g 1

;

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(a) If h(x) = f (g(x)), find h⁰(1).

(b) If H(x) = g(f (x)), find H⁰(1).

Solution:

204 ¤ CHAPTER 3 DIFFERENTIATION RULES 62.() =

4 + 3 () ⇒ ⁰() = ¹₂(4 + 3 ())^−12· 3⁰(), so

⁰(1) = ¹₂(4 + 3 (1))^−12· 3⁰(1) = ¹₂(4 + 3 · 7)^−12· 3 · 4 = ^√⁶₂₅= ⁶₅.

63. (a) () = (()) ⇒ ⁰() = ⁰(()) · ⁰(), so ⁰(1) = ⁰((1)) · ⁰(1) = ⁰(2) · 6 = 5 · 6 = 30.

(b) () = (()) ⇒ ⁰() = ⁰( ()) · ⁰(), so ⁰(1) = ⁰( (1)) · ⁰(1) = ⁰(3) · 4 = 9 · 4 = 36.

64. (a)  () = (()) ⇒ ⁰() = ⁰( ()) · ⁰(), so ⁰(2) = ⁰( (2)) · ⁰(2) = ⁰(1) · 5 = 4 · 5 = 20.

(b) () = (()) ⇒ ⁰() = ⁰(()) · ⁰(), so ⁰(3) = ⁰((3)) · ⁰(3) = ⁰(2) · 9 = 7 · 9 = 63.

65. (a) () = (()) ⇒ ⁰() = ⁰(())⁰(). So ⁰(1) = ⁰((1))⁰(1) = ⁰(3)⁰(1). To ﬁnd ⁰(3), note that  is linear from (2 4) to (6 3), so its slope is3 − 4

6 − 2= −1

4. To ﬁnd ⁰(1), note that  is linear from (0 6) to (2 0), so its slope is0 − 6

2 − 0= −3. Thus, ⁰(3)⁰(1) =

−¹4

(−3) = ³4.

(b) () = (()) ⇒ ⁰() = ⁰( ())⁰(). So ⁰(1) = ⁰( (1))⁰(1) = ⁰(2)⁰(1), which does not exist since

⁰(2)does not exist.

(c) () = (()) ⇒ ⁰() = ⁰(())⁰(). So ⁰(1) = ⁰((1))⁰(1) = ⁰(3)⁰(1). To ﬁnd ⁰(3), note that  is linear from (2 0) to (5 2), so its slope is2 − 0

5 − 2= 2

3. Thus, ⁰(3)⁰(1) =2 3

(−3) = −2.

66. (a) () = (()) ⇒ ⁰() = ⁰( ())⁰(). So ⁰(2) = ⁰( (2))⁰(2) = ⁰(1)⁰(2) ≈ (−1)(−1) = 1.

(b) () = (²) ⇒ ⁰() = ⁰(²) · 



²

= ⁰(²)(2). So ⁰(2) = ⁰(2²)(2 · 2) = 4⁰(4) ≈ 4(2) = 8.

67.The point (3 2) is on the graph of , so (3) = 2. The tangent line at (3 2) has slope∆

∆ = −4 6 = −2

3.

() =

 () ⇒ ⁰() = ¹₂[ ()]^−12· ⁰() ⇒

⁰(3) =¹₂[ (3)]^−12· ⁰(3) =¹₂(2)^−12(−²3) = − 1 3√

2 or −¹6

√2.

68. (a)  () = (^) ⇒ ⁰() = ⁰(^) 

(^) = ⁰(^)^⁻¹ (b) () = [()]^ ⇒ ⁰() =  [ ()]^⁻¹⁰()

69. (a)  () = (^) ⇒ ⁰() = ⁰(^) 

(^) = ⁰(^)^ (b) () = ^{ ()} ⇒ ⁰() = ^{ ()} 

 () = ^{ ()}⁰()

70. (a) () = ^+  () ⇒ ⁰() = ^·  + ⁰() ⇒ ⁰(0) = ⁰·  + ⁰(0) =  + 5.

⁰() = ^+ ⁰() ⇒ ⁰⁰() = ^·  + ⁰⁰() ⇒ ⁰⁰(0) = ²⁰+ ⁰⁰(0) = ²− 2.

(b) () = ^ () ⇒ ⁰() = ^⁰() +  () · ^ ⇒ ⁰(0) = ⁰⁰(0) +  (0) · ⁰= 5 + 3.

An equation of the tangent line to the graph of  at the point (0 (0)) = (0 (0)) = (0 3) is

 − 3 = (5 + 3)( − 0) or  = (5 + 3) + 3.

1

(2)

99. Let c be the x-intercept of the tangent line to the curve y = b^x(b > 0, b 6= 1) at the point (a, b^a). Show that the distance between the point (a, 0) and (c, 0) is the same for all values of a.

SECTION 3.5 Implicit Differentiation 209

99. Let c be the x-intercept of the tangent line to the curve y = b^X

(b > 0, b 共 1)at the point 仰，b^a). Show that the distance

100. On every exponential curve y = b^X(b > 0, b 手 1) ，there is exactly one point (xo, yo) at which the tangent line to the between the points 版，0)and 忙， 0)is the same for all values

of a.

curve passes through the origin. Show that in every case, yo = e. [Hint: You may wish to use Formula 1.5.10.]

y

x

101. If F ⁼ fo 9 0 h, where J, g, and h are di旺erentiablefunc- tions, use the Chain Rule to show that

F'(x) =

f'

(g(h(x)) . g'(h(x)) .h'(x)

102. If F ⁼

f

⁰g, where J and 9 are twice di叮叮entiablefunc- tions, use the Chain Rule and the Product Rule to show that the second derivative of F is given by

F 叫x) = f叫 g(x)) . [g'(X)]2 +

f'

(g(x)) . 9 叫x)

APPLIED

PROJ 的 j

WHERE SHOULD A PILOT START

0以ENT?

y=P(x)

An approach path for an aircraft landing is shown in the fìgure and satisfìes the fol1owing

之--可1_ conditions:

h

(i) The cruising altitude is h when descent starts at a horizontal distance

e

from touchdown at the origin.

(ii) The pilot must maintain a constant horizontal speed v throughout descent.

(iii) The absolute value of the vertical acceleration should not exceed a constant k (which is much less than the acceleration due to gravity).

01 e

^x 1. Find a cubic polynomial P(x) = ax³+ bx 2 + cx + d that satisfìes condition (i) by imposing suitable conditions on P(x) and P'(x) at the start of descent and at touchdown 2. Use conditions (ii) and (iii) to show that

6hv²

一一一一^- ^，，:三^L

e² ^、心

3. Suppose that an airline decides not to allow vertical acceleration of a plane to exceed k = 1385 km/h2. If the cruising altitude of a plane is 11,000 m and the speed is 480 km/h, how far away from the airport should the pilot start descent?

自 4. Graph the approach path if the conditions stated in Problem 3 are satisfìed

3.5 j lmplicit Differentiation

• Implicitly Defined Functions

The functions that we have met so far can be described by expressing one variable explic- itly in terms of another variable-for example,

y

=

#τ了。r y = x Slll X

肘， in general, y = j(x). Some functions, however, are defined implicitly by a relation Solution:

SECTION 3.4 THE CHAIN RULE ¤ 223 97. Since ^◦= 

180

rad, we have 

(sin ^◦) = 



sin₁₈₀^ 

= ₁₈₀^ cos₁₈₀^  =₁₈₀^ cos ^◦. 98. (a) () = || =√

²= (²)^12 ⇒ ⁰() = ¹₂(²)^−12(2) = √

²=  || for  6= 0.

is not differentiable at  = 0.

(b) () = |sin | =√

sin² ⇒

⁰() =¹₂(sin²)^−122 sin  cos  = sin 

|sin |cos 

=

cos  if sin   0

− cos  if sin   0

is not differentiable when  = ,  an integer.

(c) () = sin || = sin√

² ⇒

⁰() = cos || · 

||= 

||cos  =

cos  if   0

− cos  if   0

is not differentiable at 0.

99.  = ^ ⇒ ⁰= ^ln , so the slope of the tangent line to the curve  = ^at the point ( ^)is ^ln . An equation of this tangent line is then  − ^= ^ln  ( − ). If  is the intercept of this tangent line, then 0 − ^= ^ln  ( − ) ⇒

−1 = ln  ( − ) ⇒ −1

ln  =  −  ⇒ | − | =



−1 ln 



 = 1

|ln |. The distance between ( 0) and ( 0) is | − |, and this distance is the constant 1

|ln |for any . [Note: The absolute value is needed for the case 0    1 because ln  is negative there. If   1, we can write  −  = 1(ln ) as the constant distance between ( 0) and ( 0).]

100.  = ^ ⇒ ⁰= ^ln , so the slope of the tangent line to the curve  = ^at the point (0 0)is ^⁰ln . An equation of this tangent line is then  − ⁰ = ^⁰ln  ( − ⁰). Since this tangent line must pass through (0 0), we have

0 − ⁰ = ^⁰ln  (0 − ⁰), or 0= ^⁰(ln ) 0. Since (0 0)is a point on the exponential curve  = ^, we also have

0= ^⁰. Equating the expressions for 0gives ^⁰= ^⁰(ln ) 0 ⇒ 1 = (ln ) ⁰ ⇒ ⁰= 1(ln ).

So 0= ^⁰= ^⁰^{ln } [by Formula 1.5.10] = (1(ln )) ln = ¹= .

101. Let () = (()) so that  () = ((())) = (()). By the Chain Rule, we have ⁰() = ⁰(()) · ⁰()and, by the Chain Rule and substitution, we have ⁰() = ⁰(()) · ⁰() = ⁰((())) · ⁰(()) · ⁰().

102.  () =  (()) ⇒ ⁰() = ⁰(()) · ⁰()by the Chain Rule. By the Product Rule and Chain Rule we have

⁰⁰() = ⁰(()) · ⁰⁰() + ⁰() · 

⁰(()) = ⁰(()) · ⁰⁰() + ⁰() · ⁰⁰(()) ⁰()

= ⁰⁰(())[⁰()]²+ ⁰(()) · ⁰⁰()

2