Mathematics Advanced • Year 11 • Module 4 • Lesson 10

Differentiating a^x

Build procedural fluency in differentiating a^x using the ln a factor, with chain, product and quotient combinations.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the general exponential derivative:

d/dx (a^x) = ____________________

d/dx (a^kx) = ____________________ (chain rule)

Q1.2 Why does the formula d/dx (e^x) = e^x "drop" the log factor? Complete:

Because ln(____________) = ____________, so the "correction" disappears.

Q1.3 Rewrite a^x using base e. Complete:

a^x = e^{x · ____________} (since a = e^{ln a})

Stuck? Revisit lesson § Key Terms and § Concept.

2. Worked example — differentiate y = x · 5^x

Product rule, with each derivative shown. Watch where ln 5 appears.

Problem. Differentiate y = x · 5^x with respect to x.

Step 1 — Identify u and v.

u = x, v = 5^x

Reason: linear times exponential ⇒ product rule.

Step 2 — Differentiate each factor.

u' = 1, v' = 5^x ln 5

Reason: power rule on x; d/dx(a^x) = a^x ln a.

Step 3 — Apply the product rule.

dy/dx = u'v + uv' = 1 · 5^x + x · 5^x ln 5

Reason: (uv)' = u'v + uv'.

Step 4 — Factorise common 5^x.

= 5^x(1 + x ln 5)

Conclusion. dy/dx = 5^x(1 + x ln 5).

3. Faded example — fill in the missing steps

Differentiate y = 3^2x using the chain rule. 3 marks

Step 1 — Identify the inner function u (the exponent):

u = ____________ so du/dx = ____________

Step 2 — Use d/dx(a^u) = a^u · ln a · du/dx:

dy/dx = 3^2x · ln(____) · (____)

Step 3 — Simplify:

dy/dx = ____________ · 3^2x

Conclusion. dy/dx = ________________.

Stuck? Revisit lesson § Worked Example 2.

4. Graduated practice — differentiate each function

Show the ln a factor explicitly. Factorise final answers where natural.

Foundation — straight a^x and a^kx (4 questions)

Q	Function	dy/dx
4.1 1	y = 2^x
4.2 1	y = 4^x
4.3 1	y = 10^x
4.4 1	y = 2^3x

Standard — combine with product/quotient/chain (6 questions)

Name the rule used on each.

4.5 y = 7^x + 7^−x 2 marks

4.6 y = x · 3^x 2 marks

4.7 y = 5^x / x, x ≠ 0 2 marks

4.8 y = 2^x² 2 marks

4.9 y = x² · 3^x 3 marks

4.10 Find the gradient of y = 10^x at x = 0. Give an exact value. 2 marks

Extension — proof + stationary point (2 questions)

4.11 Use the identity a^x = e^{x ln a} and the chain rule to derive the formula d/dx(a^x) = a^x ln a from d/dx(e^u) = e^u · u'. 3 marks

4.12 Find the value of x where the gradient of y = (2x + 1) · 3^x is zero. Give the answer in exact form, then to 2 d.p. 4 marks

Stuck on 4.12? Apply the product rule, factor out 3^x, then set the bracket equal to 0.

5. Self-check the easy 3

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

For 4.1 I wrote d/dx(2^x) = 2^x ln 2 — and remembered the ln 2 factor (not just 2^x).

For 4.2 my answer is 4^x ln 4, with ln 4 in front (not just 4).

For 4.4 I included both the chain factor (3) and the ln a factor (ln 2), giving 3 ln 2 · 2^3x.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — General exponential derivatives

d/dx (a^x) = a^x ln a. d/dx (a^kx) = k a^kx ln a.

Q1.2 — Why e^x is special

Because ln(e) = 1, so the correction factor is 1 and d/dx(e^x) = e^x · 1 = e^x. The base e is literally chosen so this factor disappears.

Q1.3 — Rewriting a^x as base-e

a^x = e^{x · ln a}. This is the trick that derives the ln a factor (see Q4.11).

Q3 — Faded chain example y = 3^2x

Step 1: u = 2x, du/dx = 2. Step 2: dy/dx = 3^2x · ln(3) · (2). Step 3: dy/dx = 2 ln 3 · 3^2x. Conclusion: dy/dx = 2 ln 3 · 3^2x.

Q4.1 — y = 2^x

dy/dx = 2^x ln 2 ≈ 0.693 · 2^x.

Q4.2 — y = 4^x

dy/dx = 4^x ln 4 ≈ 1.386 · 4^x.

Q4.3 — y = 10^x

dy/dx = 10^x ln 10 ≈ 2.303 · 10^x.

Q4.4 — y = 2^3x

Chain rule with u = 3x, u' = 3: dy/dx = 2^3x · ln 2 · 3 = 3 ln 2 · 2^3x.

Q4.5 — y = 7^x + 7^−x

dy/dx = 7^x ln 7 + 7^−x · ln 7 · (−1) = ln 7 (7^x − 7^−x).

Q4.6 — y = x · 3^x

Product rule: u = x, v = 3^x; u' = 1, v' = 3^x ln 3. dy/dx = 1 · 3^x + x · 3^x ln 3 = 3^x(1 + x ln 3).

Q4.7 — y = 5^x / x

Quotient rule: u = 5^x, v = x; u' = 5^x ln 5, v' = 1. dy/dx = (5^x ln 5 · x − 5^x · 1) / x² = 5^x(x ln 5 − 1) / x².

Q4.8 — y = 2^x²

Chain rule with u = x², u' = 2x: dy/dx = 2^x² · ln 2 · 2x = 2x · ln 2 · 2^x².

Q4.9 — y = x² · 3^x

Product rule: u = x², v = 3^x; u' = 2x, v' = 3^x ln 3. dy/dx = 2x · 3^x + x² · 3^x ln 3 = x · 3^x(2 + x ln 3).

Q4.10 — Gradient of y = 10^x at x = 0

dy/dx = 10^x ln 10. At x = 0: dy/dx = 10⁰ · ln 10 = 1 · ln 10 = ln 10 ≈ 2.303.

Q4.11 — Derive d/dx(a^x) = a^x ln a

Write a^x = e^{x ln a} (since a = e^{ln a}).
Let u = x ln a, so u' = ln a (constant).
Chain rule: d/dx(e^u) = e^u · u' = e^{x ln a} · ln a = a^x ln a. ▮ The ln a factor is exactly the inner derivative u' = ln a.

Q4.12 — Gradient zero for y = (2x + 1) · 3^x

Product rule: u = 2x + 1, v = 3^x; u' = 2, v' = 3^x ln 3. dy/dx = 2 · 3^x + (2x + 1) · 3^x ln 3 = 3^x[2 + (2x + 1) ln 3].
Set dy/dx = 0: 3^x > 0 always, so 2 + (2x + 1) ln 3 = 0 ⇒ (2x + 1) ln 3 = −2 ⇒ 2x + 1 = −2/ln 3 ⇒ 2x = −1 − 2/ln 3 ⇒ x = (−1 − 2/ln 3) / 2 = −½ − 1/ln 3.
Numerical: x ≈ −0.5 − 1/1.0986 ≈ −0.5 − 0.9102 ≈ −1.41.

1. Quick recall

2. Worked example — differentiate y = x · 5x

3. Faded example — fill in the missing steps

4. Graduated practice — differentiate each function

Foundation — straight ax and akx (4 questions)

Standard — combine with product/quotient/chain (6 questions)

Extension — proof + stationary point (2 questions)

5. Self-check the easy 3

Q1.1 — General exponential derivatives

Q1.2 — Why ex is special

Q1.3 — Rewriting ax as base-e

Q3 — Faded chain example y = 32x

Q4.1 — y = 2x

Q4.2 — y = 4x

Q4.3 — y = 10x

Q4.4 — y = 23x

Q4.5 — y = 7x + 7−x

Q4.6 — y = x · 3x

Q4.7 — y = 5x / x

Q4.8 — y = 2x²

Q4.9 — y = x² · 3x

Q4.10 — Gradient of y = 10x at x = 0

Q4.11 — Derive d/dx(ax) = ax ln a

Q4.12 — Gradient zero for y = (2x + 1) · 3x