ln x ln 5 + x · log5e f0(x

Homework Problem Solution

§3.6 8.

f (x) = log₅(xe^x) = log₅x + log₅(e^x) = ln x

ln 5 + x · log₅e f⁰(x) = 1

ln 5 1

x + log₅e = 1 ln 5(1

x + 1)

19.

f (x) = ln(e^−x(1 + x)) = −x + ln(1 + x) f⁰(x) = −1 + 1

1 + x = −x 1 + x

25.

y⁰ = 1

x +√

1 + x² · (1 + x

√1 + x²) = (1 + x²)⁻¹² y⁰⁰ = −1

2(1 + x²)⁻³² · 2x = −x(1 + x²)⁻³²

26.

y⁰ = 1

sec x + tan x · (sec x tan x + sec²x) = sec x y⁰⁰= sec x tan x

38.

f⁰(x) = 1 ln a

6x 3x²− 2 f⁰(1) = 1

ln a 6

3 − 2 = 6 ln a , 3

⇒ ln a = 2 ⇒ a = e² 1

40.

ln y = 1

2ln x + x²+ 10 ln(x² + 1) Differentiating with respect to x gives

y⁰ y = 1

2x + 2x + 20x x²+ 1 y⁰ = y( 1

2x + 2x + 20x

x²+ 1) = √

xe^x2(x²+ 1)¹⁰( 1

2x+ 2x + 20x x²+ 1)

51.

y⁰ = 1

x²+ y²(2x + 2yy⁰) = 2x

x²+ y² + 2y x²+ y²y⁰

⇒ y⁰ = 2x x²+ x²− 2y

52. x^y = y^x ⇒ y ln x = x ln y

Differentiating both sides with respect to x:

y⁰ · ln x + y · 1

x = 1 · ln y + x ·y⁰ y

⇒ y⁰(ln x − x

y) = ln y − y x

⇒ y⁰ = ln y − ^y_x ln x −^x_y

54. y = x⁸ln x d⁹

dx⁹y = d⁸ dx⁸(dy

dx)

= d⁸

dx⁸(8x⁷ln x + x⁷)

= d⁸

dx⁸(8x⁷ln x) since d⁸

dx⁸(x⁷) = 0.

2

With this idea, we have d⁸

dx⁸(8x⁷ln x) = d⁷

dx⁷(8 · 7x⁶ln x) = · · · = d

dx(8!x⁰ln x) = 8!

x

56.

Let m = n/x, then n = xm and m → ∞ as n → ∞.

Therefore,

n→∞lim(1 + x

n)ⁿ= lim

m→∞(1 + 1

m)^mx = [ lim

m→∞(1 + 1

m)^m]^x = e^x by equation 6.

3