Solução_Calculo_Stewart_6e

F.
110 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
69. x 2 +4y 2 =5 ⇒ 2x +4(2yy 0 )=0 ⇒ y 0 = − x .Nowleth be the height of the lamp, and let (a, b) be the point of
4y
tangency of the line passing through the points (3,h) and (−5, 0). Thislinehasslope(h − 0)/[3 − (−5)] = 1 8 h.Butthe
slope of the tangent line through the point (a, b) can be expressed as y 0 = − a 4b ,oras b − 0
a − (−5) = b [since the line
a +5
passes through (−5, 0) and (a, b)], so − a 4b =
b
a +5
⇔ 4b 2 = −a 2 − 5a ⇔ a 2 +4b 2 = −5a. Buta 2 +4b 2 =5
[since (a, b) is on the ellipse], so 5=−5a ⇔ a = −1. Then4b 2 = −a 2 − 5a = −1 − 5(−1) = 4 ⇒ b =1, since the
point is on the top half of the ellipse. So h 8 =
x-axis.
b
a +5 = 1
−1+5 = 1 4
⇒
h =2. So the lamp is located 2 units above the
3.6 Derivatives of Logarithmic Functions
1. The differentiation formula for logarithmic functions,
3. f(x) =sin(lnx) ⇒ f 0 (x) =cos(lnx) ·
5. f(x) =log 2 (1 − 3x) ⇒ f 0 (x) =
d
dx (log a x) = 1 ,issimplestwhena = e because ln e =1.
x ln a
d
1 cos(ln x)
ln x =cos(lnx) · =
dx x x
1
(1 − 3x)ln2
d
−3
(1 − 3x) =
dx
(1 − 3x) ln2 or 3
(3x − 1) ln 2
7. f(x) = 5√ ln x =(lnx) 1/5 ⇒ f 0 (x) = 1 5 (ln x)−4/5 d
dx (ln x) = 1
5(ln x) 4/5 · 1
x = 1
5x 5 (ln x) 4
9. f(x) =sinx ln(5x) ⇒ f 0 (x) =sinx ·
11. F (t) =ln
F 0 (t) =3·
1
5x · d
sin x · 5
(5x)+ln(5x) · cos x = +cosx ln(5x) = sin x +cosx ln(5x)
dx 5x
x
(2t +1)3
(3t − 1) 4 =ln(2t +1)3 − ln(3t − 1) 4 =3ln(2t +1)− 4ln(3t − 1) ⇒
1
2t +1 · 2 − 4 · 1
3t − 1 · 3= 6
2t +1 − 12
3t − 1 , or combined, −6(t +3)
(2t +1)(3t − 1) .
13. g(x) =ln x √ x 2 − 1 =lnx +ln(x 2 − 1) 1/2 =lnx + 1 2 ln(x2 − 1) ⇒
g 0 (x) = 1 x + 1 2 ·
1
x 2 − 1 · 2x = 1 x + x
x 2 − 1 = x2 − 1+x · x
= 2x2 − 1
x(x 2 − 1) x(x 2 − 1)
15. f(u) =
ln u
1+ln(2u)
⇒
f 0 (u) =
[1 + ln(2u)] · 1
u − ln u · 1
2u · 2
[1 + ln(2u)] 2 =
17. y =ln 2 − x − 5x 2 ⇒ y 0 =
1
u
[1 + ln(2u) − ln u] 1+(ln2+lnu) − ln u
= =
[1 + ln(2u)] 2 u[1 + ln(2u)] 2
1
−10x − 1
· (−1 − 10x) =
2 − x − 5x2 2 − x − 5x or 10x +1
2 5x 2 + x − 2
19. y =ln(e −x + xe −x )=ln(e −x (1 + x)) = ln(e −x )+ln(1+x) =−x +ln(1+x) ⇒
y 0 = −1+ 1 −1 − x +1
= = − x
1+x 1+x 1+x
1+ln2
u[1 + ln(2u)] 2