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