Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 111
√
21. y =2x log 10 x =2x log10 x 1/2 =2x · 1 log 2 10 x = x log 10 x ⇒ y 0 1
= x ·
x ln 10 +log 10 x · 1= 1
ln 10 +log 10 x
Note:
1
ln 10 = ln e
ln 10 =log 10 e, so the answer could be written as 1
ln 10 +log 10 x =log 10 e +log 10 x =log 10 ex.
23. y = x 2 ln(2x) ⇒ y 0 = x 2 ·
y 00 =1+2x ·
1 · 2+ln(2x) · (2x) =x +2x ln(2x)
2x ⇒
1 · 2+ln(2x) · 2=1+2+2ln(2x) =3+2ln(2x)
2x
25. y =ln x + √ 1+x 2 ⇒
y 0 1
=
x + √ 1+x 2
=
1
x + √ 1+
1+x 2
d √
x + 1+x
2
=
dx
x
√ = 1+x
2
1
x + √ 1+ 1
1+x (1 + 2 2 x2 ) −1/2 (2x)
1
x + √ 1+x 2 ·
√
1+x2 + x
√
1+x
2
=
1
√
1+x
2
⇒
y 00 = − 1 2 (1 + x2 ) −3/2 (2x) =
−x
(1 + x 2 ) 3/2
27. f(x) =
x
1 − ln(x − 1)
⇒
−1
[1 − ln(x − 1)] · 1 − x ·
f 0 (x) =
x − 1
[1 − ln(x − 1)] 2 =
=
2x − 1 − (x − 1) ln(x − 1)
(x − 1)[1 − ln(x − 1)] 2
(x − 1)[1 − ln(x − 1)] + x
x − 1
x − 1 − (x − 1) ln(x − 1) + x
=
[1 − ln(x − 1)] 2 (x − 1)[1 − ln(x − 1)] 2
Dom(f) ={x | x − 1 > 0 and 1 − ln(x − 1) 6= 0} = {x | x>1 and ln(x − 1) 6= 1}
= x | x>1 and x − 1 6=e 1 = {x | x>1 and x 6= 1+e} =(1, 1+e) ∪ (1 + e, ∞)
29. f(x) =ln(x 2 − 2x) ⇒ f 0 (x) =
1
2(x − 1)
(2x − 2) =
x 2 − 2x x(x − 2) .
Dom(f) ={x | x(x − 2) > 0} =(−∞, 0) ∪ (2, ∞).
31. f(x) = ln x ⇒ f 0 (x) = x2 (1/x) − (ln x)(2x) x − 2x ln x x(1 − 2lnx)
= = = 1 − 2lnx ,
x 2 (x 2 ) 2 x 4
x 4
x 3
so f 0 (1) = 1 − 2ln1 = 1 − 2 · 0 =1.
1 3 1
33. y =ln xe x2 =lnx +lne x2 =lnx + x 2 ⇒ y 0 = 1 +2x. At(1, 1), the slope of the tangent line is
x
y 0 (1) = 1 + 2 = 3, and an equation of the tangent line is y − 1=3(x − 1),ory =3x − 2.
35. f(x) =sinx +lnx ⇒ f 0 (x) =cosx +1/x.
This is reasonable, because the graph shows that f increases when f 0 is
positive, and f 0 (x) =0when f has a horizontal tangent.