Mathematics Advanced • Year 11 • Module 4 • Lesson 11

Differentiating ln x

Build procedural fluency in differentiating natural logarithm functions, including chain, product and quotient combinations.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete each derivative:

d/dx ( ln x ) = ____________ (for x > 0)

d/dx ( ln g(x) ) = ____________ (chain rule form)

Q1.2 State the domain on which y = ln x is defined: ____________

Q1.3 Use a log law to rewrite each expression as a sum or difference of logarithms (so it is easier to differentiate):

ln(3x) = ____________________________

ln(x²) = ____________________________

Stuck? Revisit lesson § Key terms — "Derivative of ln x".

2. Worked example — differentiate y = ln(x² + 1)

Follow each line. Every step has a reason on the right.

Problem. Differentiate y = ln(x² + 1).

Step 1 — Identify the inside function.

g(x) = x² + 1 ⇒ g'(x) = 2x

Reason: chain-rule form needs g'(x) on top and g(x) on the bottom.

Step 2 — Apply the chain rule for ln.

dy/dx = g'(x) / g(x) = 2x / (x² + 1)

Reason: d/dx ( ln g(x) ) = g'(x) / g(x).

Step 3 — Domain check.

x² + 1 > 0 for all real x ✓

Reason: the argument of ln must be positive; here it always is.

Conclusion. dy/dx = 2x / (x² + 1), valid for all real x.

3. Faded example — fill in the missing steps

Differentiate y = x ln x. Fill in each blank, then state the stationary x-value. 4 marks

Step 1 — Identify u and v for the product rule:

u = ____________ , v = ____________

Step 2 — Differentiate each factor:

u' = ____________ , v' = ____________

Step 3 — Apply the product rule dy/dx = u'v + uv':

dy/dx = ________ · ln x + ________ · ________ = ln x + ________

Step 4 — Set dy/dx = 0 to find the stationary point:

ln x + 1 = 0 ⇒ ln x = ________ ⇒ x = ________

Conclusion. The stationary point of y = x ln x is at x = ____________.

Stuck? Revisit lesson § Worked example 3 — y = x ln x.

4. Graduated practice — differentiate each function

For each function, find dy/dx. Show at least one line of working for Standard and Extension questions.

Foundation — direct rule (4 questions)

Q	Function	dy/dx
4.1 1	y = ln x
4.2 1	y = ln(4x)
4.3 1	y = ln(x⁵)
4.4 1	y = 3 ln x

Standard — chain, product, quotient (6 questions)

Use the form g'(x)/g(x) for chain-rule items; state u, v, u', v' for product/quotient items.

4.5 y = ln(5x + 2) 2 marks

4.6 y = ln(x² + 4) 2 marks

4.7 y = x² ln x 2 marks

4.8 y = (ln x) / x 2 marks

4.9 y = (x + 1) ln x 2 marks

4.10 Find the gradient of y = ln(x² + 4) at x = 2. 2 marks

Extension — combine ideas (2 questions)

4.11 By first using log laws, differentiate y = ln( x² (x + 1) ). Verify your answer simplifies to 2/x + 1/(x+1). 3 marks

4.12 Show that y = ln x and y = x have only one point at which they share the same gradient, and find the x-value of that point. 3 marks

Stuck on 4.12? Equal gradients means setting d/dx (ln x) = d/dx (x).

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I wrote 1/x (not 1/ln x and not ln x).

For 4.2 I either used the chain rule (4 / 4x = 1/x) or first split ln(4x) = ln 4 + ln x — both give 1/x.

For 4.3 I either used the chain rule (5x⁴ / x⁵ = 5/x) or first split ln(x⁵) = 5 ln x — both give 5/x.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Derivative rules

d/dx ( ln x ) = 1/x. d/dx ( ln g(x) ) = g'(x) / g(x).

Q1.2 — Domain

y = ln x is defined for x > 0. (For x < 0 use ln |x|, which has the same derivative 1/x.)

Q1.3 — Log laws

ln(3x) = ln 3 + ln x. ln(x²) = 2 ln x. (Splitting first turns a chain-rule job into single-term derivatives.)

Q3 — Faded example y = x ln x

Step 1: u = x, v = ln x. Step 2: u' = 1, v' = 1/x.
Step 3: dy/dx = 1 · ln x + x · 1/x = ln x + 1.
Step 4: ln x = −1 ⇒ x = e⁻¹ = 1/e.
Stationary point at x = 1/e ≈ 0.368.

Q4.1 — y = ln x

dy/dx = 1/x.

Q4.2 — y = ln(4x)

Chain rule: dy/dx = 4 / (4x) = 1/x. Or split: ln(4x) = ln 4 + ln x, derivative = 0 + 1/x = 1/x.

Q4.3 — y = ln(x⁵)

Split first: ln(x⁵) = 5 ln x, so dy/dx = 5/x. (Chain-rule check: 5x⁴ / x⁵ = 5/x.)

Q4.4 — y = 3 ln x

dy/dx = 3 · (1/x) = 3/x.

Q4.5 — y = ln(5x + 2)

g = 5x + 2, g' = 5. dy/dx = 5 / (5x + 2). Valid for 5x + 2 > 0, i.e. x > −2/5.

Q4.6 — y = ln(x² + 4)

g = x² + 4, g' = 2x. dy/dx = 2x / (x² + 4). Valid for all real x (denominator always > 0).

Q4.7 — y = x² ln x

Product rule: u = x², v = ln x, u' = 2x, v' = 1/x. dy/dx = 2x · ln x + x² · (1/x) = 2x ln x + x = x(2 ln x + 1).

Q4.8 — y = (ln x) / x

Quotient rule: u = ln x, v = x, u' = 1/x, v' = 1. dy/dx = [(1/x)·x − ln x · 1] / x² = (1 − ln x) / x².

Q4.9 — y = (x + 1) ln x

Product rule: u = x + 1, v = ln x, u' = 1, v' = 1/x. dy/dx = 1 · ln x + (x + 1) · (1/x) = ln x + 1 + 1/x.

Q4.10 — Gradient of y = ln(x² + 4) at x = 2

From 4.6, dy/dx = 2x / (x² + 4). At x = 2: dy/dx = 4 / 8 = 1/2.

Q4.11 — y = ln( x² (x + 1) )

Split using log laws: y = ln(x²) + ln(x + 1) = 2 ln x + ln(x + 1).
dy/dx = 2 · (1/x) + 1/(x + 1) = 2/x + 1/(x + 1). ✓ As claimed.

Q4.12 — Where ln x and x share a gradient

Gradients: d/dx (ln x) = 1/x; d/dx (x) = 1. Setting 1/x = 1 gives x = 1. This is the only solution because 1/x is strictly decreasing on x > 0, so it crosses the horizontal line y = 1 exactly once.