Solução_Calculo_Stewart_6e

F.
100 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
19. y =(2x − 5) 4 (8x 2 − 5) −3 ⇒
y 0 =4(2x − 5) 3 (2)(8x 2 − 5) −3 +(2x − 5) 4 (−3)(8x 2 − 5) −4 (16x)
=8(2x − 5) 3 (8x 2 − 5) −3 − 48x(2x − 5) 4 (8x 2 − 5) −4
[This simplifies to 8(2x − 5) 3 (8x 2 − 5) −4 (−4x 2 +30x − 5).]
21. y =
x 2 3
+1
x 2 − 1
⇒
x
y 0 2 2
+1
=3
·
x 2 − 1
x 2 +1
=3
x 2 − 1
d
dx
x 2 +1
=3
x 2 − 1
23. y = e x cos x ⇒ y 0 = e x cos x ·
25. F (z) =
2
· 2x[x2 − 1 − (x 2 +1)]
(x 2 − 1) 2 =3
1/2
z − 1 z − 1
z +1 = z +1
F 0 (z) = 1 −1/2 z − 1
·
2 z +1
d
dz
x 2 2
+1
· (x2 − 1)(2x) − (x 2 +1)(2x)
x 2 − 1
(x 2 − 1) 2
x 2 2
+1
·
x 2 − 1
2x(−2)
(x 2 − 1) = −12x(x2 +1) 2
2 (x 2 − 1) 4
d
dx (x cos x) =ex cos x [x(− sin x)+(cosx) · 1] = e x cos x (cos x − x sin x)
⇒
z − 1
= 1 z +1 2
1/2 z +1
·
z − 1
(z +1)(1)− (z − 1)(1)
(z +1) 2
= 1 (z +1) 1/2 z +1− z +1
· = 1 (z +1) 1/2
2 (z − 1)
1/2
(z +1) 2 2 (z − 1) · 2
1/2 (z +1) = 1
2 (z − 1) 1/2 (z +1) 3/2
27. y =
r
√
r2 +1
⇒
y 0 =
√
√
r 2
r2 +1(1)− r · 1
2 (r2 +1) −1/2 r2 +1− √
(2r)
√
r2 +1 r2 +1
2
= √
r2 +1 2
=
√
r2 +1 √ r 2 +1− r 2
√
r2 +1
√
r2 +1 2
=
r 2 +1 − r 2
√
r2 +1 1
3
=
(r 2 +1) or 3/2 (r2 +1) −3/2
Another solution: Write y as a product and make use of the Product Rule. y = r(r 2 +1) −1/2
⇒
y 0 = r · − 1 2 (r2 +1) −3/2 (2r)+(r 2 +1) −1/2 · 1=(r 2 +1) −3/2 [−r 2 +(r 2 +1) 1 ]=(r 2 +1) −3/2 (1) = (r 2 +1) −3/2 .
The step that students usually have trouble with is factoring out (r 2 +1) −3/2 . But this is no different than factoring out x 2
from x 2 + x 5 ; that is, we are just factoring out a factor with the smallest exponent that appears on it. In this case, − 3 2 is
smaller than − 1 2 .
29. y = sin(tan 2x) ⇒ y 0 =cos(tan2x) ·
31. Using Formula 5 and the Chain Rule, y =2 sin πx ⇒
y 0 =2 sin πx (ln 2) ·
d
dx (tan 2x) = cos(tan 2x) · d
sec2 (2x) ·
dx (2x) = 2 cos(tan 2x)sec2 (2x)
d
dx (sin πx) =2sin πx (ln 2) · cos πx · π =2 sin πx (π ln 2) cos πx