Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ¤ 97
13. y = sin x ⇒ y 0 = x2 cos x − (sin x)(2x) x(x cos x − 2sinx) x cos x − 2sinx
x 2 (x 2 ) 2 = =
x 4
x 3
15. Using Exercise 3.2.55(a), f(x) =xe x csc x ⇒
17.
19.
f 0 (x)=(x) 0 e x csc x + x(e x ) 0 csc x + xe x (csc x) 0 =1e x csc x + xe x csc x + xe x (− cot x csc x)
= e x csc x (1 + x − x cot x)
d
d 1 (sin x)(0) − 1(cos x)
(csc x) = =
dx dx sin x
sin 2 = − cos x
x
sin 2 x = − 1
sin x · cos x = − csc x cot x
sin x
d
d
(cot x) =
dx dx
cos x
(sin x)(− sin x) − (cos x)(cos x)
=
sin x
sin 2 = − sin2 x +cos 2 x
x
sin 2 = − 1
x sin 2 x = − csc2 x
21. y =secx ⇒ y 0 =secx tan x,soy 0 ( π 3 )=secπ 3 tan π 3 =2√ 3. An equation of the tangent line to the curve y =secx
at the point π
, 2 is y − 2=2 √ 3 √ √
x − π 3 3 or y =2 3 x +2−
2
3 3 π.
23. y = x +cosx ⇒ y 0 =1− sin x. At(0, 1), y 0 =1, and an equation of the tangent line is y − 1=1(x − 0),ory = x +1.
25. (a) y =2x sin x ⇒ y 0 =2(x cos x +sinx · 1). At π
2 ,π ,
y 0 =2 π
cos π
2 2 +sinπ 2 =2(0+1)=2, and an equation of the
tangent line is y − π =2
x − π 2 ,ory =2x.
(b)
27. (a) f(x) =secx − x ⇒ f 0 (x) =secx tan x − 1
(b)
Note that f 0 =0where f has a minimum. Also note that f 0 is negative
when f is decreasing and f 0 is positive when f is increasing.
29. H(θ) =θ sin θ ⇒ H 0 (θ) =θ (cos θ) +(sinθ) · 1=θ cos θ +sinθ ⇒
H 00 (θ) =θ (− sin θ)+(cosθ) · 1+cosθ = −θ sin θ +2cosθ
31. (a) f(x) = tan x − 1
sec x
⇒
f 0 (x) = sec x(sec2 x) − (tan x − 1)(sec x tan x)
(sec x) 2
(b) f(x) = tan x − 1
sec x
=
sin x
cos x − 1
=
1
cos x
(c) From part (a), f 0 (x) = 1+tanx
sec x
sin x − cos x
cos x
1
cos x
= sec x(sec2 x − tan 2 x +tanx)
sec 2 x
= 1+tanx
sec x
=sinx − cos x ⇒ f 0 (x) =cosx − (− sin x) =cosx +sinx
= 1
sec x + tan x
sec x =cosx +sinx, which is the expression for f 0 (x) in part (b).