Mathematics Advanced • Year 11 • Module 4 • Lesson 10

Differentiating a^x

Past-paper-style problems on the ln a factor, chain rule for general bases, and a proof-based extended response.

Master · Past-Paper Style

1. Short-answer questions

1.1 Differentiate y = 7^x + 7^−x, simplifying your answer with ln 7 factored out. 2 marks Band 3

1.2 Differentiate y = 2^x², showing the chain-rule step. 3 marks Band 4

1.3 Find the value of x at which the gradient of y = (2x + 1) · 3^x is zero. Give the answer in exact form (in terms of ln 3) and to 2 d.p. 4 marks Band 5

Stuck on 1.3? Product rule, factor out 3^x, set the bracket to 0.

2. Extended response

2.1 (a) Use the identity a^x = e^{x ln a} and the chain rule to prove that d/dx(a^x) = a^x ln a for any a > 0.
(b) Use part (a) to deduce d/dx(a^kx) for any constant k.
(c) The number of confirmed cases in an outbreak is modelled by N(t) = 100 · 2^t/3, where t is days from the start.
    (i) Differentiate N(t) and write dN/dt in the form k · N(t) for an explicit constant k.
    (ii) Compare k with the "natural" rate λ from the equivalent base-e model N(t) = 100 e^λt. State λ exactly.
    (iii) Find the value of t at which dN/dt = 200 cases per day. Give your answer to the nearest day.
   7 marks    Band 5-6

Explicit marking criteria

Part (a) — 2 marks

• 1 mark — writes a^x = e^{x ln a} and identifies inner function u = x ln a, with u' = ln a (constant).

• 1 mark — applies the chain rule on e^u: d/dx(e^{x ln a}) = e^{x ln a} · ln a = a^x ln a.

Part (b) — 1 mark

• 1 mark — applies the chain rule with inner u = kx, u' = k: d/dx(a^kx) = a^kx · ln a · k = k a^kx ln a.

Part (c)(i) — 1 mark

• 1 mark — dN/dt = 100 · (1/3) · 2^t/3 · ln 2 = (ln 2 / 3) · N(t).

Part (c)(ii) — 1 mark

• 1 mark — λ = ln 2 / 3 (the same constant; this is just the rewriting 2^t/3 = e^{(t/3) ln 2}).

Part (c)(iii) — 2 marks

• 1 mark — sets up dN/dt = 200: (ln 2 / 3) · 100 · 2^t/3 = 200, so 2^t/3 = 6/ln 2.

• 1 mark — solves for t = 3 log₂(6/ln 2) ≈ 3 × 3.113 ≈ 9.34, so t ≈ 9 days.

Your response:

Stuck on (c)(iii)? Take log₂ of both sides after isolating 2^t/3.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — y = 7^x + 7^−x (2 marks)

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

Marking notes. 1 mark — d/dx(7^x) = 7^x ln 7. 1 mark — chain-rule factor −1 on the second term and ln 7 factored out. Missing the ln 7 factor anywhere = 1/2 maximum.

1.2 — y = 2^x² (3 marks)

Sample response. Let u = x², u' = 2x. Using d/dx(a^u) = a^u · ln a · u':
dy/dx = 2^x² · ln 2 · 2x = 2x · ln 2 · 2^x².

Marking notes. 1 mark — identifies u = x² and u' = 2x. 1 mark — applies d/dx(a^u) = a^u · ln a · u' explicitly. 1 mark — final expression includes all three factors (2x, ln 2, 2^x²). Missing ln 2 = 1/3 maximum.

1.3 — Zero gradient for y = (2x + 1) · 3^x (4 marks)

Sample response. 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, so 2 + (2x + 1) ln 3 = 0 ⇒ (2x + 1) ln 3 = −2 ⇒ 2x + 1 = −2/ln 3 ⇒ x = (−1 − 2/ln 3) / 2 = −½ − 1/ln 3.
Numerical: 1/ln 3 ≈ 1/1.0986 ≈ 0.9102, so x ≈ −0.5 − 0.910 ≈ −1.41.

Marking notes. 1 mark — product rule applied with correct u', v' (including ln 3 factor on v'). 1 mark — factor out 3^x. 1 mark — solve for x in exact form. 1 mark — numerical to 2 d.p. Common error: students forget the ln 3 factor on v' and conclude x = −3/2 — costs 2 marks.

2.1 — Extended response (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

Part (a). Since a = e^{ln a}, raising both sides to power x gives

a^x = (e^{ln a})^x = e^{x ln a}. [1 mark — identity stated; inner function u = x ln a with u' = ln a.]

Differentiate using d/dx(e^u) = e^u · u':

d/dx(a^x) = d/dx(e^{x ln a}) = e^{x ln a} · ln a = a^x · ln a. ▮ [1 mark.]

Part (b). For a^kx, let u = kx, so u' = k. Chain rule on the result of (a) (or rewrite as e^{kx ln a}):

d/dx(a^kx) = a^kx · ln a · k = k a^kx ln a. [1 mark.]

Part (c)(i). N(t) = 100 · 2^t/3, so k = 1/3 and a = 2. By part (b),

dN/dt = (1/3) · 100 · 2^t/3 · ln 2 = (ln 2 / 3) · [100 · 2^t/3] = (ln 2 / 3) · N(t). [1 mark.]

Part (c)(ii). Rewrite 2^t/3 = e^{(t/3) ln 2} = e^{(ln 2 / 3) t}. So N(t) = 100 e^λt with

λ = ln 2 / 3 ≈ 0.231 per day. [1 mark.] This is the same constant as k in (i) — the two model forms are just different ways of writing the same growth.

Part (c)(iii). Set dN/dt = 200:

(ln 2 / 3) · 100 · 2^t/3 = 200 ⇒ 2^t/3 = 200 · 3 / (100 ln 2) = 6 / ln 2 ≈ 6 / 0.6931 ≈ 8.656. [1 mark — set-up.]

Take log₂: t/3 = log₂(8.656) = ln(8.656) / ln 2 ≈ 2.158 / 0.6931 ≈ 3.114. t ≈ 9.34 ⇒ t ≈ 9 days. [1 mark.]

Total: 7/7.

Band descriptors for marker.

Band 3: States a^x = e^{x ln a} but does not differentiate; gives part (c)(i) constant ≈ 0.23 without exact form. ≈ 2-3 marks.

Band 4: Proof in (a) complete; (b) follows but stated without chain-rule detail; (c)(i) and (c)(ii) correct; (c)(iii) numerical only. ≈ 4-5 marks.

Band 5: Proofs in (a) and (b) clean; (c)(i)-(ii) recognise the equivalence; (c)(iii) algebra correct but log-conversion not shown. ≈ 5-6 marks.

Band 6: All algebra precise; (a) and (b) include the "u' = ln a" / "u' = k" step explicitly; (c)(ii) names λ = ln 2 / 3 and notes the equivalence to k from (i); (c)(iii) shows the log₂ step and rounds correctly. 7/7.