3.6 SOLUTIONS
2.
4.
6.
8.
10.
dy
= 2 · 3(8x − 1)2 · 8 = 48(8x − 1)2
dx
dy
= − sin(− x3 )(− 13 ) = 31 sin(− x3 )
dx
dy
= cos(x − cos x)(1 + sin x)
dx
dy
= − sec(x2 + 7x) tan(x2 + 7x)(2x + 7)
dx
y = u9 , where u = 4 − 3x. Then
dy
= 9(4 − 3x)8 (−3) = −27(4 − 3x)8
dx
√
14. y = u, where u = 3x2 − 4x + 6. Then
dy
1
= (3x2 − 4x + 6)−1/2 (6x − 4).
dx
2
20. y = eu , where u = 32 x. Then
2 2x
dy
= e3.
dx
3
√
u
2
22. y = e , where u = 4 x + x . Then
√
dy
2
= e4 x+x (2x−1/2 + 2x)
dx
ds
= 3π
26.
cos( 3π
t) −
2
2
dt
36. Use product rule
3π
2
sin( 3π
t)
2
dy
= 2e−2x − 2(1 + 2x)e−2x .
dx
38. Use product rule
dy
3
3
= (18x − 6)ex + (9x2 − 6x + 2)(ex )(3x2 ).
dx
dy
= 2 sec(πt) sec(πt) cos(πt)π = 2π sec2 (πt) cos(πt)
dt
dy
= 3(esin(t/2) )2 esin(t/2) cos( 21 t) 12 = 32 (esin(t/2) )2 esin(t/2) cos( 21 t)
58.
dt
dy
62.
= − sin(5 sin( 13 t))(5 cos( 13 t))( 13 ) = − 53 sin(5 sin( 31 t)) cos( 13 t)
dx
68. We compute
52.
dy
= 4 cos3 (sec2 (3t))(− sin(sec2 (3t))(2 sec(3t) sec(3t) tan(3t)3)
dt
= −24 cos3 (sec2 (3t)) sin(sec2 (3t)) sec(3t) sec(3t) tan(3t).
1
2
3.6 SOLUTIONS
dy
du
= u−2 . Furthermore
= (1 − x)−2 . When x = −1, we
du
dx
have u = 1/2. Then by the chain rule
dy dy du 1
=
= 4 · = 1.
dx
du
dx
4
80. Let y = f (u). Then
x=−1
u=1/2
x=−1
dy
du
82. Let y = f (u). Then
= 1 − cos23 u and
= π. When x = 1/4, u = π/4. By chain
du
dx
rule
dy du 2
dy 8
=
= (1 − √
)π = (1 − √ )π.
3
dx x=1/4 du u=π/4 dx x=1/4
( 2/2)
2
88.
a. We compute
5f 0 (1) − g 0 (1) = 5(−1/3) − (−8/3) = 1.
b. Use product rule
f 0 (0)g 3 (0) + f (0)3g 2 (0)g 0 (0) = 5(13 ) + 1(3(12 )(1/3)) = 6.
c. Use quotient rule
(−4 + 1)(−1/3) − (3)(−8/3)
(g(1) + 1)f 0 (1) − f (1)g 0 (1)
=
= 1.
2
(g(1) + 1)
(−4 + 1)2
d. Use chain rule
1
f 0 (g(0))g 0 (0) = f 0 (1)(1/3) = (−1/3)(1/3) = − .
9
e. Use chain rule
40
g 0 (f (0))f 0 (0) = g 0 (1)5 = (−8/3)5 = − .
3
f. Use power rule with chain rule
1
−2(111 + f (1))−3 (11(110 ) + f 0 (1)) = −2(4)−3 (32/3) = − .
3
g. We compute
20
f 0 (0 + g(0))(1 + g 0 (0)) = f 0 (1)(1 + 1/3) = 5(1 + 1/3) = .
3
dy
92. Note that this exercise is allowing you to verify that
= 32 x1/2 .
dx
dy
du
1 −1/2
2
a.
= 3u and
= 2x
. Then by chain rule, we have
du
dx
√ 1
dy
3
= 3( x)2 x−1/2 = x1/2 .
dx
2
2
dy
du
b.
= 12 u1/2 and
= 3x2 . Then by the chain rule
du
dx
dy
1
3
3
= (x3 )−1/2 (3x2 ) = x−3/2 x2 = x1/2 .
dx
2
2
2