Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 3.4 THE CHAIN RULE ¤ 101
33. y =sec 2 x +tan 2 x =(secx) 2 +(tanx) 2 ⇒
y 0 =2(secx)(sec x tan x) + 2(tan x)(sec 2 x)=2sec 2 x tan x +2sec 2 x tan x =4sec 2 x tan x
1 − e
2x
35. y =cos
1+e 2x
⇒
1 − e
y 0 2x
= − sin
·
1+e 2x
d 1 − e
2x
1 − e
2x
= − sin
· (1 + e2x )(−2e 2x ) − (1 − e 2x )(2e 2x )
dx 1+e 2x 1+e 2x (1 + e 2x ) 2
1 − e
2x
1 − e
2x
= − sin
· −2e2x (1 + e 2x )+(1− e 2x )
= − sin
1+e 2x (1 + e 2x ) 2
1+e 2x
· −2e2x (2)
(1 + e 2x ) = 4e 2x 1 − e
2x
2 (1 + e 2x ) · sin 2 1+e 2x
37. y =cot 2 (sin θ) =[cot(sinθ)] 2 ⇒
y 0 = 2[cot(sin θ)] ·
d
dθ [cot(sin θ)] = 2 cot(sin θ) · [− csc2 (sin θ) · cos θ] =−2cosθ cot(sin θ) csc 2 (sin θ)
39. f(t) =tan(e t )+e tan t ⇒ f 0 (t) =sec 2 (e t ) · d
dt (et )+e tan t · d
dt (tan t) =sec2 (e t ) · e t + e tan t · sec 2 t
41. f(t) =sin 2 e sin2 t
= sin e
t 2 sin2 ⇒
f 0 (t)=2 sin e sin2 t
· d
=2sin e sin2 t
cos
dt sin
e sin2 t
e sin2 t
=4sin e sin2 t
cos e sin2 t
e sin2t sin t cos t
=2sin e sin2 t
· cos e sin2 t
· d
· e sin2t · d
dt sin2 t =2sin e sin2 t
cos
dt esin2 t
e sin2 t
e sin2t · 2sint cos t
43. g(x) =(2ra rx + n) p ⇒
g 0 (x) =p(2ra rx + n) p−1 ·
45. y =cos sin(tan πx) = cos(sin(tan πx)) 1/2 ⇒
y 0 = − sin(sin(tan πx)) 1/2 ·
= − sin sin(tan πx)
2 sin(tan πx)
d
dx (2rarx + n) =p(2ra rx + n) p−1 · 2ra rx (ln a) · r =2r 2 p(ln a)(2ra rx + n) p−1 a rx
d
dx (sin(tan πx))1/2 = − sin(sin(tan πx)) 1/2 · 1 (sin(tan d
2 πx))−1/2 · (sin(tan πx))
dx
· cos(tan πx) ·
= −π cos(tan πx)sec2 (πx)sin sin(tan πx)
2 sin(tan πx)
d
dx tan πx = − sin sin(tan πx)
2 · cos(tan πx) · sec 2 (πx) · π
sin(tan πx)
47. h(x) = √ x 2 +1 ⇒ h 0 (x) = 1 2 (x2 +1) −1/2 (2x) =
x
√
x2 +1
⇒
h 00 (x) =
√
1
x2 +1· 1 − x
2 (x2 +1) −1/2 (2x)
√
x2 +1 2
=
x 2 +1 −1/2 (x 2 +1)− x 2
(x 2 +1) 1 =
1
(x 2 +1) 3/2