Solução_Calculo_Stewart_6e

F.
104 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
83. By the Chain Rule, a(t) = dv
dt = dv ds
ds dt = dv
dv
v(t) =v(t)
ds ds .
Thederivativedv/dt is the rate of change of the velocity
with respect to time (in other words, the acceleration) whereas the derivative dv/ds is the rate of change of the velocity with
respect to the displacement.
85. (a) Using a calculator or CAS, we obtain the model Q = ab t with a ≈ 100.0124369 and b ≈ 0.000045145933.
(b) Use Q 0 (t) =ab t ln b (from Formula 5) with the values of a and b from part (a) to get Q 0 (0.04) ≈−670.63 μA.
The result of Example 2 in Section 2.1 was −670 μA.
87. (a) Derive gives g 0 (t) =
45(t − 2)8
without simplifying. With either Maple or Mathematica, we first get
(2t +1)
10
g 0 (t − 2)8 (t − 2)9
(t) =9 − 18 , and the simplification command results in the expression given by Derive.
(2t +1)
9
(2t +1)
10
(b) Derive gives y 0 =2(x 3 − x +1) 3 (2x +1) 4 (17x 3 +6x 2 − 9x +3)without simplifying. With either Maple or
Mathematica, we first get y 0 =10(2x +1) 4 (x 3 − x +1) 4 +4(2x +1) 5 (x 3 − x +1) 3 (3x 2 − 1). Ifweuse
Mathematica’s Factor or Simplify,orMaple’sfactor, we get the above expression, but Maple’s simplify gives
the polynomial expansion instead. For locating horizontal tangents, the factored form is the most helpful.
89. (a) If f is even, then f(x) =f(−x). Using the Chain Rule to differentiate this equation, we get
f 0 (x) =f 0 (−x) d
dx (−x) =−f 0 (−x). Thus,f 0 (−x) =−f 0 (x),sof 0 is odd.
(b) If f is odd, then f(x) =−f(−x).
even.
91. (a)
(b)
d
dx (sinn x cos nx) =n sin n−1 x cos x cos nx +sin n x (−n sin nx)
Differentiating this equation, we get f 0 (x) =−f 0 (−x)(−1) = f 0 (−x),sof 0 is
[Product Rule]
= n sin n−1 x (cos nx cos x − sin nx sin x) [factor out n sin n−1 x]
= n sin n−1 x cos(nx + x) [Addition Formula for cosine]
= n sin n−1 x cos[(n +1)x] [factor out x]
d
dx (cosn x cos nx) =n cos n−1 x (− sin x)cosnx +cos n x (−n sin nx)
[Product Rule]
= −n cos n−1 x (cos nx sin x +sinnx cos x) [factor out −n cos n−1 x]
= −n cos n−1 x sin(nx + x) [Addition Formula for sine]
= −n cos n−1 x sin[(n +1)x] [factor out x]
93. Since θ ◦ =
π
d
180 θ rad, we have
dθ (sin θ◦ )= d
sin
π
180
dθ
θ = π cos π θ = π cos 180 180 180 θ◦ .
95. The Chain Rule says that dy
dx = dy
du
dy
d 2 y
dx = d
2 dx
dx
du
dx ,so
= d
dx
dy du
du dx
d
=
dx
dy du
du
dx + dy d
du dx
du
dx
[Product Rule]
d
=
du
dy
du
du du
dx dx + dy
d 2 u
du dx = d2 y du
2 du 2 dx
2
+ dy d 2 u
du dx 2