Sec 4 A-Math · Differentiation of trigonometric, exponential and logarithmic functions

Differentiating logarithmic functions questions with answers

Practice questions on differentiating logarithmic functions, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

Given that $y = \ln(3x + 7)$ , find $\dfrac{dy}{dx}$ .

Step 1 of 3

Let u be what the logarithm is taken of, and differentiate it

u = 3x + 7

\dfrac{du}{dx} = 3

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Questions

Question 1ln(ax + b)
Given that $y = 2\ln(4x - 3)$ , find $\dfrac{dy}{dx}$ .
Show the worked solution
1. Step 1: Let u be what the logarithm is taken of, and differentiate it $u = 4x - 3$ $\dfrac{du}{dx} = 4$
2. Step 2: Differentiate ln u with respect to u $\dfrac{dy}{du} = \dfrac{2}{u}$
3. Step 3: Chain rule: multiply, then write u in terms of x again $\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$ $\dfrac{dy}{dx} = \dfrac{2}{4x - 3} \times 4$ $= \dfrac{8}{4x - 3}$
Question 2ln(ax + b)
Given that $y = \ln(7x + 3)$ , find $\dfrac{dy}{dx}$ .
Show the worked solution
1. Step 1: Let u be what the logarithm is taken of, and differentiate it $u = 7x + 3$ $\dfrac{du}{dx} = 7$
2. Step 2: Differentiate ln u with respect to u $\dfrac{dy}{du} = \dfrac{1}{u}$
3. Step 3: Chain rule: multiply, then write u in terms of x again $\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$ $\dfrac{dy}{dx} = \dfrac{1}{7x + 3} \times 7$ $= \dfrac{7}{7x + 3}$
Question 3Laws of logarithms first
Given that $y = \ln\sqrt{2x + 1}$ , find $\dfrac{dy}{dx}$ .
Show the worked solution
1. Step 1: Write the square root as a power, then use a law of logarithms $y = \dfrac{1}{2}\ln(2x + 1)$
2. Step 2: Let u be what the logarithm is taken of, and differentiate it $u = 2x + 1$ $\dfrac{du}{dx} = 2$
3. Step 3: Differentiate ln u with respect to u $\dfrac{dy}{du} = \dfrac{1}{2u}$
4. Step 4: Chain rule: multiply, then write u in terms of x again $\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$ $\dfrac{dy}{dx} = \dfrac{1}{2(2x + 1)} \times 2$ $= \dfrac{1}{2x + 1}$
Question 4Laws of logarithms first
Given that $y = \ln\left(\dfrac{x + 5}{3x + 1}\right)$ , find $\dfrac{dy}{dx}$ .
Show the worked solution
1. Step 1: Use a law of logarithms to split the logarithm first $y = \ln(x + 5) - \ln(3x + 1)$
2. Step 2: Differentiate each logarithm: the derivative of ln(ax + b) is a ÷ (ax + b) $\dfrac{dy}{dx} = \dfrac{1}{x + 5} - \dfrac{3}{3x + 1}$
Question 5Chain, product and quotient rules
Given that $y = 4\ln(x^{2} + 7)$ , find $\dfrac{dy}{dx}$ .
Show the worked solution
1. Step 1: Let u be what the logarithm is taken of, and differentiate it $u = x^{2} + 7$ $\dfrac{du}{dx} = 2x$
2. Step 2: Differentiate ln u with respect to u $\dfrac{dy}{du} = \dfrac{4}{u}$
3. Step 3: Chain rule: multiply, then write u in terms of x again $\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$ $\dfrac{dy}{dx} = \dfrac{4}{x^{2} + 7} \times 2x$ $= \dfrac{8x}{x^{2} + 7}$
Question 6Chain, product and quotient rules
Given that $y = 2\ln(x^{2} + 7)$ , find $\dfrac{dy}{dx}$ .
Show the worked solution
1. Step 1: Let u be what the logarithm is taken of, and differentiate it $u = x^{2} + 7$ $\dfrac{du}{dx} = 2x$
2. Step 2: Differentiate ln u with respect to u $\dfrac{dy}{du} = \dfrac{2}{u}$
3. Step 3: Chain rule: multiply, then write u in terms of x again $\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$ $\dfrac{dy}{dx} = \dfrac{2}{x^{2} + 7} \times 2x$ $= \dfrac{4x}{x^{2} + 7}$

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Questions

Question 1ln(ax + b)

Given that $y = 2\ln(4x - 3)$ , find $\dfrac{dy}{dx}$ .

Show the worked solution

Step 1: Let u be what the logarithm is taken of, and differentiate it $u = 4x - 3$ $\dfrac{du}{dx} = 4$
Step 2: Differentiate ln u with respect to u $\dfrac{dy}{du} = \dfrac{2}{u}$
Step 3: Chain rule: multiply, then write u in terms of x again $\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$ $\dfrac{dy}{dx} = \dfrac{2}{4x - 3} \times 4$ $= \dfrac{8}{4x - 3}$

Question 2ln(ax + b)

Given that $y = \ln(7x + 3)$ , find $\dfrac{dy}{dx}$ .

Show the worked solution

Step 1: Let u be what the logarithm is taken of, and differentiate it $u = 7x + 3$ $\dfrac{du}{dx} = 7$
Step 2: Differentiate ln u with respect to u $\dfrac{dy}{du} = \dfrac{1}{u}$
Step 3: Chain rule: multiply, then write u in terms of x again $\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$ $\dfrac{dy}{dx} = \dfrac{1}{7x + 3} \times 7$ $= \dfrac{7}{7x + 3}$

Question 3Laws of logarithms first

Given that $y = \ln\sqrt{2x + 1}$ , find $\dfrac{dy}{dx}$ .

Show the worked solution

Step 1: Write the square root as a power, then use a law of logarithms $y = \dfrac{1}{2}\ln(2x + 1)$
Step 2: Let u be what the logarithm is taken of, and differentiate it $u = 2x + 1$ $\dfrac{du}{dx} = 2$
Step 3: Differentiate ln u with respect to u $\dfrac{dy}{du} = \dfrac{1}{2u}$
Step 4: Chain rule: multiply, then write u in terms of x again $\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$ $\dfrac{dy}{dx} = \dfrac{1}{2(2x + 1)} \times 2$ $= \dfrac{1}{2x + 1}$

Question 4Laws of logarithms first

Given that $y = \ln\left(\dfrac{x + 5}{3x + 1}\right)$ , find $\dfrac{dy}{dx}$ .

Show the worked solution

Step 1: Use a law of logarithms to split the logarithm first $y = \ln(x + 5) - \ln(3x + 1)$
Step 2: Differentiate each logarithm: the derivative of ln(ax + b) is a ÷ (ax + b) $\dfrac{dy}{dx} = \dfrac{1}{x + 5} - \dfrac{3}{3x + 1}$

Question 5Chain, product and quotient rules

Given that $y = 4\ln(x^{2} + 7)$ , find $\dfrac{dy}{dx}$ .

Show the worked solution

Step 1: Let u be what the logarithm is taken of, and differentiate it $u = x^{2} + 7$ $\dfrac{du}{dx} = 2x$
Step 2: Differentiate ln u with respect to u $\dfrac{dy}{du} = \dfrac{4}{u}$
Step 3: Chain rule: multiply, then write u in terms of x again $\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$ $\dfrac{dy}{dx} = \dfrac{4}{x^{2} + 7} \times 2x$ $= \dfrac{8x}{x^{2} + 7}$

Question 6Chain, product and quotient rules

Given that $y = 2\ln(x^{2} + 7)$ , find $\dfrac{dy}{dx}$ .

Show the worked solution

Step 1: Let u be what the logarithm is taken of, and differentiate it $u = x^{2} + 7$ $\dfrac{du}{dx} = 2x$
Step 2: Differentiate ln u with respect to u $\dfrac{dy}{du} = \dfrac{2}{u}$
Step 3: Chain rule: multiply, then write u in terms of x again $\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}$ $\dfrac{dy}{dx} = \dfrac{2}{x^{2} + 7} \times 2x$ $= \dfrac{4x}{x^{2} + 7}$