A-Level Maths differentiation revision: rules, applications, worked
A-Level differentiation runs from first principles through the chain, product and quotient rules to parametric and implicit differentiation, then applies all of it to tangents and normals, stationary points, and connected rates of change. The chain rule does most of the work — most exam questions are applications of it in disguise — and the commonest lost marks are skipped steps in 'show that' questions and dy/dx assembled upside down in parametric work.
Differentiation is the friendlier half of calculus — the rules are mechanical and the same three applications recur every year. The marks go to students who are automatic on the mechanics, because the exam's difficulty is all in the disguise: a connected-rates question is just the chain rule wearing a context.
The specification, in plain English
| Technique | What it looks like | Watch for |
|---|---|---|
| First principles | lim(h→0) [f(x+h) − f(x)]/h | Usually a 'show that' for x² or x³ — write every step |
| Chain rule | y = (3x²+1)⁵, y = e^(sin x) | The outer/inner split; missing the inner derivative |
| Product rule | y = x²·e^x, y = x·ln x | Sign errors expanding (uv′ + vu′) |
| Quotient rule | y = (2x+1)/(x−3) | The squared denominator; simplify after |
| Parametric | x = t², y = 3t → dy/dx = (dy/dt)/(dx/dt) | Upside-down assembly; find t before substituting |
| Implicit | x² + y² = 25, x·y = e^y | d/dx of y-terms brings a dy/dx; collect them after |
The applications that carry the marks
- Tangents and normals — gradient of tangent is dy/dx at the point; the normal's gradient is the negative reciprocal.
- Stationary points — solve dy/dx = 0, then classify with d²y/dx² (positive → minimum, negative → maximum) or a sign test.
- Connected rates of change — dh/dt = dh/dV × dV/dt; build the chain so the unwanted variable cancels.
- Small increments — δy ≈ (dy/dx)·δx, a quiet two-marker that appears regularly.
- Increasing and decreasing functions — dy/dx > 0 means strictly increasing; examiners ask you to prove it.
Worked questions
Differentiate y = x²·e^(3x).
- Product rule: u = x², v = e^(3x).
- u′ = 2x. For v, chain rule: v′ = 3e^(3x).
- dy/dx = u′v + uv′ = 2x·e^(3x) + x²·3e^(3x).
- Factor out e^(3x): dy/dx = e^(3x)(3x² + 2x) = x·e^(3x)(3x + 2).
- Factored form is what the mark scheme shows — an unfactored correct answer still scores, but factorising makes the next part ('hence find stationary points') visible.
dy/dx = x·e^(3x)(3x + 2).
A curve is defined by x = 2t − t², y = t² + 4t. Find dy/dx in terms of t, and the coordinates of the stationary point.
- dx/dt = 2 − 2t and dy/dt = 2t + 4.
- dy/dx = (dy/dt)/(dx/dt) = (2t + 4)/(2 − 2t) = (t + 2)/(1 − t).
- Stationary when dy/dx = 0 — the numerator is zero: t = −2.
- Coordinates: x = 2(−2) − 4 = −8, y = 4 − 8 = −4.
- Note dx/dt ≠ 0 there, so it's a genuine stationary point, not a vertical tangent.
dy/dx = (t + 2)/(1 − t); stationary point (−8, −4) at t = −2.
Water drains from a cone (V = ⅓πr²h, with r = h/2) at 20 cm³/s. Find dh/dt when h = 10.
- Eliminate r using the cone's proportions: r = h/2 gives V = ⅓π(h²/4)h = πh³/12.
- Differentiate: dV/dh = πh²/4.
- Build the chain: dh/dt = dh/dV × dV/dt = (1 ÷ dV/dh) × (−20).
- At h = 10: dV/dh = 25π, so dh/dt = −20/(25π) = −4/(5π) ≈ −0.255.
- The negative sign is the answer's meaning — the level is falling. Examiners award a mark for stating it correctly signed.
dh/dt ≈ −0.255 cm/s — the water level falls about 2.5 mm each second at that instant.
Where marks get dropped
- Forgetting the inner derivative in the chain rule — e^(3x) differentiates to 3e^(3x), not e^(3x).
- dy/dx upside down in parametric questions — it is dy/dt over dx/dt, always.
- Implicit: differentiating y² to 2y and forgetting the dy/dx that follows it.
- Solving dy/dx = 0 then not classifying the point — the question said 'determine the nature'.
- First-principles 'show that' compressed into one line — the limit steps are the marks.
- Normal gradient as the same as the tangent's — it's the negative reciprocal.
Your differentiation checklist
- I can differentiate products, quotients and chains mixed together without hesitating.
- I can do a parametric question end-to-end including the tangent equation.
- I can differentiate an implicit equation and isolate dy/dx.
- I can set up a connected-rates chain from a worded problem, with units.
- I can find and classify stationary points using the second derivative.
- I can do a first-principles proof for x² or 1/x line by line.
If your differentiation is mechanically fine but the applications — rates, tangents, 'show that's — keep slipping, the gap is setup, not technique. That is exactly the gap a worked board explanation fixes: Lumi draws the chain, shows which rate feeds which, and then gives you one to set up yourself.
What to take from this
- Chain rule is the workhorse — dy/dx = dy/du × du/dx underpins parametric, implicit and rates questions.
- In parametric differentiation, dy/dx = (dy/dt) ÷ (dx/dt). Upside down is the classic error.
- Stationary points: dy/dx = 0 finds them; d²y/dx² or a sign test classifies them.
- Connected rates of change are just the chain rule with units — build the chain from the rates you're given.
- Differentiate to check an integration, integrate to sanity-check a differentiated answer — the exam rewards self-checkers.
Questions people also ask
Parametric is for curves given as x = f(t), y = g(t) — you differentiate each with respect to t and divide. Implicit is for relations like x² + y² = 25 where y isn't isolated — you differentiate term by term and every y-term picks up a dy/dx.
Yes — 'show from first principles' questions appear regularly and are pure method marks. Practise the proofs for x², x³ and 1/x until the limit manipulation is automatic.
The chain rule says dh/dt = dh/dV × dV/dt — the dV cancels. The skill is writing down the chain that links the rate you want to the rate you are given, then differentiating to fill in the middle link.
Second derivative is faster when it exists and is easy — positive means minimum, negative maximum. Use the sign test (checking dy/dx either side) when the second derivative is zero, undefined, or ugly to compute.