Mathematics Advanced • Year 11 • Module 4 • Lesson 9

Differentiating e^x

Build procedural fluency in differentiating e^x, e^kx, e^u(x), and combinations using the product and quotient rules.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the three core derivatives:

d/dx (e^x) = ____________

d/dx (e^kx) = ____________

d/dx (e^u(x)) = ____________ (chain rule)

Q1.2 What makes e^x unique among all functions? Complete in one sentence:

e^x is the only function equal to its own ____________________.

Q1.3 Write "true" or "false":

A. d/dx (e^2x) = e^2x ⇒ ____________

B. e^x > 0 for every real x ⇒ ____________

Stuck? Revisit lesson § Key Terms and § Concept.

2. Worked example — differentiate y = x² e^x

Two functions multiplied ⇒ product rule. Each line names which factor we're differentiating.

Problem. Differentiate y = x² e^x with respect to x.

Step 1 — Identify u and v.

u = x², v = e^x

Reason: two functions of x multiplied together.

Step 2 — Differentiate each factor.

u' = 2x, v' = e^x

Reason: power rule on x²; e^x is its own derivative.

Step 3 — Apply the product rule.

dy/dx = u'v + uv' = 2x · e^x + x² · e^x

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

Step 4 — Factorise common e^x and x.

= e^x(2x + x²) = x e^x(2 + x)

Reason: factorise to make later steps (stationary points etc.) easier.

Conclusion. dy/dx = x e^x(x + 2).

3. Faded example — fill in the missing steps

Differentiate y = e^x² using the chain rule. 3 marks

Step 1 — Identify u (the exponent):

u = ____________

Step 2 — Differentiate u:

du/dx = ____________

Step 3 — Apply chain rule:

dy/dx = e^u · du/dx = e^____ · ____________

Conclusion. dy/dx = ____________________.

Stuck? d/dx (e^u) = e^u · u'.

4. Graduated practice — differentiate each function

Where possible, factorise your final answer.

Foundation — e^kx directly (4 questions)

Q	Function	dy/dx
4.1 1	y = e^5x
4.2 1	y = e^−2x
4.3 1	y = 3 e^x
4.4 1	y = e^4x + e^−x

Standard — product, quotient, chain (6 questions)

Name the rule used (product / quotient / chain) on each.

4.5 y = x e^x 2 marks

4.6 y = e^2x / x, x ≠ 0 2 marks

4.7 y = e^{3x + 1} 2 marks

4.8 y = x³ e^2x 3 marks

4.9 y = e^{x² + 1} 2 marks

4.10 Find the gradient of y = e^x² at x = 1. 2 marks

Extension — stationary points (2 questions)

4.11 Find the stationary point of y = x e^−x and determine its nature using the second-derivative test. 4 marks

4.12 Show that y = A e^kx satisfies dy/dx = ky for any constants A and k. 2 marks

Stuck on 4.11? Use the product rule to differentiate, set dy/dx = 0, then check d²y/dx² at the critical x.

5. Self-check the easy 3

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

For 4.1 (e^5x) I used the chain rule with k = 5, getting 5 e^5x.

For 4.2 (e^−2x) I correctly carried the negative sign through: −2 e^−2x.

For 4.3 (3 e^x) the constant 3 stays in front: 3 e^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 — Core derivatives

d/dx (e^x) = e^x. d/dx (e^kx) = k e^kx. d/dx (e^u(x)) = e^u(x) · u'(x).

Q1.2 — Unique property of e^x

e^x is the only function equal to its own derivative.

Q1.3 — True/false

A. False — d/dx (e^2x) = 2 e^2x, not e^2x. The chain-rule factor k = 2 must appear. B. True — exponentials are always strictly positive.

Q3 — Faded chain-rule example

Step 1: u = x². Step 2: du/dx = 2x. Step 3: dy/dx = e^x² · 2x = 2x e^x².

Q4.1 — y = e^5x

dy/dx = 5 e^5x.

Q4.2 — y = e^−2x

dy/dx = −2 e^−2x.

Q4.3 — y = 3 e^x

dy/dx = 3 e^x. The constant multiple stays.

Q4.4 — y = e^4x + e^−x

dy/dx = 4 e^4x − e^−x. Differentiate term by term.

Q4.5 — y = x e^x

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

Q4.6 — y = e^2x / x

Quotient rule with u = e^2x, v = x: u' = 2 e^2x, v' = 1. dy/dx = (2 e^2x · x − e^2x · 1)/x² = e^2x(2x − 1)/x².

Q4.7 — y = e^{3x + 1}

Chain rule with u = 3x + 1, u' = 3. dy/dx = 3 e^{3x + 1}.

Q4.8 — y = x³ e^2x

Product rule with u = x³, v = e^2x: u' = 3x², v' = 2 e^2x. dy/dx = 3x² e^2x + x³ · 2 e^2x = x² e^2x(3 + 2x).

Q4.9 — y = e^{x² + 1}

Chain rule with u = x² + 1, u' = 2x. dy/dx = 2x e^{x² + 1}.

Q4.10 — Gradient of y = e^x² at x = 1

dy/dx = 2x e^x². At x = 1: gradient = 2(1) e¹ = 2e ≈ 5.437.

Q4.11 — Stationary point of y = x e^−x

Product rule: dy/dx = 1 · e^−x + x · (−e^−x) = e^−x(1 − x). Set dy/dx = 0: e^−x > 0 always, so 1 − x = 0 ⇒ x = 1. y(1) = 1 · e⁻¹ = 1/e. Stationary point at (1, 1/e).
Second derivative: d²y/dx² = −e^−x(1 − x) + e^−x(−1) = e^−x(x − 2). At x = 1: e⁻¹(−1) = −1/e < 0. Local maximum.

Q4.12 — A e^kx satisfies dy/dx = ky

Differentiate using d/dx(e^kx) = k e^kx: dy/dx = A · k e^kx = k (A e^kx) = k y. ▮ This is why exponentials are the natural solutions to the differential equation dy/dx = ky (proportional growth/decay).

1. Quick recall

2. Worked example — differentiate y = x² ex

3. Faded example — fill in the missing steps

4. Graduated practice — differentiate each function

Foundation — ekx directly (4 questions)

Standard — product, quotient, chain (6 questions)

Extension — stationary points (2 questions)

5. Self-check the easy 3

Q1.1 — Core derivatives

Q1.2 — Unique property of ex

Q1.3 — True/false

Q3 — Faded chain-rule example

Q4.1 — y = e5x

Q4.2 — y = e−2x

Q4.3 — y = 3 ex

Q4.4 — y = e4x + e−x

Q4.5 — y = x ex

Q4.6 — y = e2x / x

Q4.7 — y = e3x + 1

Q4.8 — y = x³ e2x

Q4.9 — y = ex² + 1

Q4.10 — Gradient of y = ex² at x = 1

Q4.11 — Stationary point of y = x e−x

Q4.12 — A ekx satisfies dy/dx = ky