A Level Maths 9709 · Paper 3 Pure Mathematics 3 · 3.4 (also Paper 2) · ~6 min read

Further differentiation

Derivatives of eˣ, ln x, sin x, cos x, tan x and tan⁻¹x, the chain, product and quotient rules together, implicit differentiation and parametric differentiation, with tangents, normals and stationary points.

Derivatives of eˣ & ln xTrig derivativestan⁻¹xChain ruleProduct ruleQuotient ruleImplicit differentiationParametric differentiation

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Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

Six standard derivatives: eˣ → eˣ; ln x → 1/x; sin x → cos x; cos x → −sin x; tan x → sec²x; tan⁻¹x → 1/(1 + x²). Angles are in radians.
Product rule: (uv)' = u'v + uv'. Quotient rule: (u/v)' = (u'v − uv')/v². Chain rule: multiply by the derivative of the inside.
Implicit differentiation: differentiate every term with respect to x; each y-term gets a factor dy/dx (chain rule).
Parametric: dy/dx = (dy/dt) ÷ (dx/dt).

By the end you can

Differentiate exponential, logarithmic and trigonometric functions, including tan⁻¹x, and composites of them.
Use the product and quotient rules.
Find dy/dx for curves defined implicitly or parametrically.
Find tangents, normals and stationary points for these curves.

1Standard derivatives

The list

f(x)	f'(x)	With the chain rule
eˣ	eˣ	e^{ax + b} → a e^{ax + b}; e^g(x) → g'(x)e^g(x)
ln x	1/x	ln(ax + b) → a/(ax + b); ln g(x) → g'(x)/g(x)
sin x	cos x	sin(ax + b) → a cos(ax + b)
cos x	−sin x	cos(ax + b) → −a sin(ax + b)
tan x	sec²x	tan(ax) → a sec²(ax)
tan⁻¹x	1/(1 + x²)	tan⁻¹(ax) → a/(1 + a²x²)

All of these are in the formula list FORMULA LIST (MF19). The skill is combining them with the rules.
Why d/dx tan⁻¹x = 1/(1 + x²): if y = tan⁻¹x then tan y = x. Differentiating implicitly: sec²y × dy/dx = 1, so dy/dx = 1/sec²y = 1/(1 + tan²y) = 1/(1 + x²).
Logs first: simplify ln(x³(2x + 1)) = 3 ln x + ln(2x + 1) before differentiating; it's quicker and safer.

Worked example

Differentiate (a) e^3x² (b) ln(2x + 1) (c) sin²x (d) tan⁻¹(2x).

(a) 6x e^3x² B1
(b) 2/(2x + 1) B1
(c) 2 sin x cos x (= sin 2x) B1
(d) 2/(1 + 4x²) B1

Common mistakeUsing double-angle formulae to differentiate sin 2x. Just use the chain rule: 2 cos 2x. The 2024 Paper 3 report advises against unnecessary identities.

2Product and quotient rules

The rules

d/dx(uv) = u dv/dx + v du/dx FORMULA LIST (MF19)

product rule

d/dx(u/v) = (v du/dx − u dv/dx)/v² FORMULA LIST (MF19)

quotient rule

Worked example

Find the x-coordinate of the stationary point of y = x² ln x, and its y-coordinate exactly.

dy/dx = 2x ln x + x² × (1/x) = 2x ln x + x M1 A1
x(2 ln x + 1) = 0; x > 0, so ln x = −½ M1
x = e^−1/2, y = e⁻¹ × (−½) = −1/(2e) A1

Worked example

Differentiate y = (x² + 1)/(x − 2).

dy/dx = [2x(x − 2) − (x² + 1)(1)]/(x − 2)² M1 A1
= (x² − 4x − 1)/(x − 2)² A1

Worked example

Find the x-coordinate of the stationary point of y = e^2x sin x for 0 < x < π.

dy/dx = 2e^2x sin x + e^2x cos x = e^2x(2 sin x + cos x) M1 A1
e^2x ≠ 0 ⇒ tan x = −½ M1
in (0, π): x = π − 0.4636 = 2.68 A1

Common mistakeQuotient-rule sign errors (v u' − u v', in that order) and missing the chain rule inside, e.g. for e^{sin x} or cos²x. Both were common in 2024.

3Implicit differentiation

Every term, with dy/dx on the y-terms

d/dx(y²) = 2y dy/dx; d/dx(xy) = y + x dy/dx (product rule); d/dx(e^y) = e^y dy/dx; d/dx(ln y) = (1/y) dy/dx.
Then collect the dy/dx terms on one side and factorise.

Implicit differentiation of a circle gives the tangent gradient at any point.

Worked example

The curve x² + 3xy + y² = 11 passes through (1, 2). Find the equation of the tangent there.

2x + 3y + 3x dy/dx + 2y dy/dx = 0 M1 A1
dy/dx = −(2x + 3y)/(3x + 2y) = −8/7 at (1, 2) M1
y − 2 = −(8/7)(x − 1) ⇒ 8x + 7y = 22 A1

Common mistakeForgetting the product rule on xy or on terms like x ln y. The 2024 Paper 2 and 3 reports both flagged ln-terms not differentiated as products.

4Parametric differentiation

dy/dx = (dy/dt)/(dx/dt)

Each value of t gives a point; the gradient is the ratio of the rates of change.

Worked example

A curve has parametric equations x = 3 cos t, y = 2 sin t. Find the gradient of the normal at t = π/4.

dx/dt = −3 sin t, dy/dt = 2 cos t B1
dy/dx = 2 cos t/(−3 sin t) = −(2/3) cot t = −2/3 at t = π/4 M1 A1
normal gradient = 3/2 A1

Finding a point from a condition: for example, the tangent is horizontal when dy/dt = 0 (and dx/dt ≠ 0); vertical when dx/dt = 0.
Converting to Cartesian form (eliminating t) is sometimes asked: here cos t = x/3 and sin t = y/2, so x²/9 + y²/4 = 1.

Linking questionPaper 1 differentiation is in Differentiation; these derivatives reverse into the integrals of Further integration; and the iteration in Numerical solution of equations often starts from a stationary point equation.

Test yourself — cover the answers first

Quick check

Differentiate e^−4x.
Differentiate ln(x² + 5).
Differentiate tan 3x.
Find d/dx(xy) implicitly.
x = t², y = t³. Find dy/dx.

Exam-style

Find the coordinates of the stationary point of y = x e^−2x and show it is a maximum.
The curve x³ + y³ = 9xy passes through (2, 4). Find dy/dx at this point.
Show that the derivative of tan⁻¹(x/2) is 2/(4 + x²).

Show answers

−4e^−4x
2x/(x² + 5)
3 sec²3x
y + x dy/dx
3t/2
dy/dx = e^−2x(1 − 2x) = 0 ⇒ x = ½, y = 1/(2e); d²y/dx² = e^−2x(4x − 4) < 0 at x = ½
3x² + 3y² dy/dx = 9y + 9x dy/dx ⇒ dy/dx = (3y − x²)/(y² − 3x) = 8/10 = 0.8
chain rule: (1/2) × 1/(1 + x²/4) = (1/2) × 4/(4 + x²) = 2/(4 + x²)

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