Solução_Calculo_Stewart_6e

F.
96 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
We will sometimes use the form f 0 g + fg 0 rather than the form fg 0 + gf 0 for the Product Rule.
55. (a) (fgh) 0 =[(fg)h] 0 =(fg) 0 h +(fg)h 0 =(f 0 g + fg 0 )h +(fg)h 0 = f 0 gh + fg 0 h + fgh 0
(b) Putting f = g = h in part (a), we have d
dx [f(x)]3 =(fff) 0 = f 0 ff + ff 0 f + fff 0 =3fff 0 =3[f(x)] 2 f 0 (x).
(c)
d
dx (e3x )= d
dx (ex ) 3 =3(e x ) 2 e x =3e 2x e x =3e 3x
57. For f(x) =x 2 e x , f 0 (x) =x 2 e x + e x (2x) =e x (x 2 +2x). Similarly, we have
f 00 (x) =e x (x 2 +4x +2)
f 000 (x) =e x (x 2 +6x +6)
f (4) (x) =e x (x 2 +8x + 12)
f (5) (x) =e x (x 2 +10x +20)
It appears that the coefficient of x in the quadratic term increases by 2 with each differentiation. The pattern for the
constant terms seems to be 0=1· 0, 2=2· 1, 6=3· 2, 12 = 4 · 3, 20 = 5 · 4. So a reasonable guess is that
f (n) (x) =e x [x 2 +2nx + n(n − 1)].
Proof: Let S n be the statement that f (n) (x) =e x [x 2 +2nx + n(n − 1)].
1. S 1 is true because f 0 (x) =e x (x 2 +2x).
2. Assume that S k is true; that is, f (k) (x) =e x [x 2 +2kx + k(k − 1)]. Then
f (k+1) (x) = d
f (k) (x) = e x (2x +2k)+[x 2 +2kx + k(k − 1)]e x
dx
= e x [x 2 +(2k +2)x +(k 2 + k)] = e x [x 2 +2(k +1)x +(k +1)k]
This shows that S k+1 is true.
3. Therefore, by mathematical induction, S n is true for all n;thatis,f (n) (x) =e x [x 2 +2nx + n(n − 1)] for every
positive integer n.
3.3 Derivatives of Trigonometric Functions
1. f(x) =3x 2 − 2cosx ⇒ f 0 (x) =6x − 2(− sin x) =6x +2sinx
3. f(x) =sinx + 1 2 cot x ⇒ f 0 (x) =cosx − 1 2 csc2 x
5. g(t) =t 3 cos t ⇒ g 0 (t) =t 3 (− sin t)+(cost) · 3t 2 =3t 2 cos t − t 3 sin t or t 2 (3 cos t − t sin t)
7. h(θ) =cscθ + e θ cot θ ⇒ h 0 (θ) =− csc θ cot θ + e θ (− csc 2 θ)+(cotθ)e θ = − csc θ cot θ + e θ (cot θ − csc 2 θ)
9. y =
x
2 − tan x ⇒ y0 = (2 − tan x)(1) − x(− sec2 x)
= 2 − tan x + x sec2 x
(2 − tan x) 2 (2 − tan x) 2
11. f(θ) = sec θ
1+secθ
⇒
f 0 (θ) =
(1 + sec θ)(sec θ tan θ) − (sec θ)(sec θ tan θ) (sec θ tan θ) [(1+secθ) − sec θ] sec θ tan θ
= =
(1 + sec θ) 2 (1 + sec θ) 2 (1 + sec θ) 2