Solução_Calculo_Stewart_6e

F.
94 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
35. (a) y = f(x) = 1
1+x 2 ⇒
f 0 (x) = (1 + x2 )(0) − 1(2x)
= −2x . So the slope of the
(1 + x 2 ) 2 (1 + x 2 )
2
(b)
tangent line at the point
−1, 1 2 is f 0 (−1) = 2 2 = 1 2 2
and its
equation is y − 1 2 = 1 2 (x +1)or y = 1 2 x +1.
37. (a) f(x) = ex
⇒ f 0 (x) = x3 (e x ) − e x (3x 2 )
= x2 e x (x − 3)
= ex (x − 3)
x 3 (x 3 ) 2 x 6 x 4
(b)
f 0 =0when f has a horizontal tangent line, f 0 is negative when
f is decreasing, and f 0 is positive when f is increasing.
39. (a) f(x) =(x − 1)e x ⇒ f 0 (x) =(x − 1)e x + e x (1) = e x (x − 1+1)=xe x .
f 00 (x) =x(e x )+e x (1) = e x (x +1)
(b)
f 0 =0when f has a horizontal tangent and f 00 =0when f 0 has a
horizontal tangent. f 0 is negative when f is decreasing and positive when f
is increasing. f 00 is negative when f 0 is decreasing and positive when f 0 is
increasing. f 00 is negative when f is concave down and positive when f is
concave up.
41. f(x) = x2
1+x ⇒ f 0 (x) = (1 + x)(2x) − x2 (1)
(1 + x) 2 = 2x +2x2 − x 2
(1 + x) 2 = x2 +2x
x 2 +2x +1
⇒
so f 00 (1) =
f 00 (x) = (x2 +2x +1)(2x +2)− (x 2 +2x)(2x +2)
(x 2 +2x +1) 2 = (2x +2)(x2 +2x +1− x 2 − 2x)
[(x +1) 2 ] 2
=
2
(1 + 1) 3 = 2 8 = 1 4 .
2(x +1)(1) 2
=
(x +1) 4 (x +1) 3 ,
43. We are given that f(5) = 1, f 0 (5) = 6, g(5) = −3,andg 0 (5) = 2.
(a) (fg) 0 (5) = f(5)g 0 (5) + g(5)f 0 (5) = (1)(2) + (−3)(6) = 2 − 18 = −16
0 f
(b) (5) = g(5)f 0 (5) − f(5)g 0 (5) (−3)(6) − (1)(2)
= = − 20 g
[g(5)] 2 (−3) 2 9
0 g
(c) (5) = f(5)g0 (5) − g(5)f 0 (5)
=
f
[f(5)] 2
(1)(2) − (−3)(6)
(1) 2 =20
45. f(x) =e x g(x) ⇒ f 0 (x) =e x g 0 (x)+g(x)e x = e x [g 0 (x)+g(x)]. f 0 (0) = e 0 [g 0 (0) + g(0)] = 1(5 + 2) = 7