Lesson 8: Derivatives of Logarithmic and Exponential Functions (worksheet solutions)
Solutions to Worksheet for Sections 3.1–3.3
Derivatives of Exponential and Logarithmic Functions
V63.0121, Calculus I
Summer 2010
Find the derivatives of the following functions.
2
−3x
1. y = e2x
Solution. We have
dy 2 d 2
= e2x −3x 2×2 − 3x = e2x −3x (4x − 3)
dx dx
2. y = 62x
Solution. We have
y = (ln 6) · 62x · 2 = (2 ln 6)62x
3. y = ln(x3 + 9)
Solution. We have
dy 1 d 3×2
= 3 x3 + 9 = 3
dx x + 9 dx x +9
4. y = log3 ex
Solution. By brute force we have
1 1
y = · ex =
(ln 3) · ex ln 3
But slightly more elegantly, we notice that
ln ex x
log3 ex = =
ln 3 ln 3
dy 1
So = makes a lot of sense.
dx ln 3
1
5. y = log10 3θ2 −θ
Solution.
dy 1 2 −1/2 1 2
= log10 3θ −θ · θ 2 −θ
· (ln 3) · 3θ −θ (2θ − 1)
dx 2 (ln 10)3
ln 3 2θ − 1
= ·
2 ln 10 log10 3θ2 −θ
There’s some simplifications we could do before differentiation, however.
2
−θ
ln 3θ θ2 − θ ln 3 ln 3 2
θ 2 −θ
log10 3 = = = θ −θ
ln 10 ln 10 ln 10
So
ln 3
y= · θ2 − θ
ln 10
ln 3 2θ − 1
y = · √
ln 10 2 θ2 − θ
6. y = sin2 x + 2sin x
Solution.
y = 2 sin x cos x + 2sin x · ln 2 · cos x
Use logarithmic differentiation to find the derivatives of the following functions.
7. y = x x2 − 1
Solution. We have
1
ln(x2 − 1)
ln y = ln x +
2
1 dy 1 1 2x
= + · 2
y dx x 2 x −1
dy 1 x
= x x2 − 1 + 2
dx x x −1
8. y = (x − 1)(x − 2)(x − 3)
2
Solution. We have
1
ln y = (ln(x − 1) + ln(x − 2) + ln(x − 3))
2
1 dy 1 1 1 1
= + +
y dx 2 x−1 x−2 x−3
dy 1 1 1 1
= (x − 1)(x − 2)(x − 3) + +
dx 2 x−1 x−2 x−3
x(x − 1)3/2
9. y = √
x+1
Solution. We have
3 1
ln y = ln x +ln(x − 1) − ln(x + 1)
2 2
1 dy 1 3 1 1 1
= + · − ·
y dx x 2 x−1 2 x+1
dy x(x − 1)3/2 1 3 1 1 1
= √ + · − ·
dx x+1 x 2 x−1 2 x+1
3
