Cambridge A Level Mathematics 9709 · Pure 1

Differentiation

Differentiation finds the gradient function: from $y = f(x)$ it produces $\tfrac{dy}{dx}$ , the rate at which $y$ changes with $x$ . In Cambridge 9709 Pure 1 it powers four question types — tangents and normals, stationary points, increasing/decreasing functions, and rates of change — and together they are consistently among the largest mark blocks on the paper.

Updated 20 August 2026

The power rule

Multiply by the power, then knock the power down by one:

y = ax^n \;\Rightarrow\; \frac{dy}{dx} = nax^{n-1}

It works for every rational power — negative and fractional included — which is why P1 questions arrive dressed in roots and fractions: rewrite first, then differentiate.

y = 3\sqrt{x} - \frac{2}{x^2} = 3x^{1/2} - 2x^{-2} \;\Rightarrow\; \frac{dy}{dx} = \tfrac{3}{2}x^{-1/2} + 4x^{-3}

Constants vanish; sums differentiate term by term. Most P1 slips on this rule happen in the rewriting, not the calculus.

The chain rule

For a function of a function, differentiate the outside, keep the inside, multiply by the inside’s derivative:

y = (2x^2-5)^6 \;\Rightarrow\; \frac{dy}{dx} = 6(2x^2-5)^5 \times 4x

In P1 the chain rule appears wherever a bracket carries a power — including negative powers like $y = (3x+1)^{-2}$ — and in connected rates of change (below). The factor from the inside (here $4x$ ) is the mark examiners look for.

Tangents and normals

The derivative evaluated at a point is the gradient of the tangent there. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal:

m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}

Differentiate.
Substitute the $x$ -coordinate to get a numerical gradient.
Pick the right gradient for what was asked — tangent uses $m$ , normal uses $-1/m$ .
Line through the point: $y - y_1 = m(x - x_1)$ .

Stationary points and their nature

Stationary points solve $\tfrac{dy}{dx} = 0$ . Their nature comes from the second derivative:

\frac{d^2y}{dx^2} > 0 \;\Rightarrow\; \text{minimum} \qquad \frac{d^2y}{dx^2} < 0 \;\Rightarrow\; \text{maximum}

A function is increasing where $\tfrac{dy}{dx} > 0$ and decreasing where $\tfrac{dy}{dx} < 0$ — P1 asks this as “find the set of values of $x$ for which $f$ is decreasing”, which is a quadratic inequality in disguise.

Worked example

The curve y = x³ − 6x² + 9x + 2. Find the stationary points and determine their nature.

Differentiate: $\tfrac{dy}{dx} = 3x^2 - 12x + 9$ .

Solve = 0 (with working — factorise): $3(x^2-4x+3) = 3(x-1)(x-3) = 0$ , so $x = 1$ or $x = 3$ . Substituting back: $(1, 6)$ and $(3, 2)$ .

Nature: $\tfrac{d^2y}{dx^2} = 6x - 12$ . At $x=1$ : $-6 < 0$ → maximum at $(1,6)$ . At $x=3$ : $+6 > 0$ → minimum at $(3,2)$ .

Note the shape of the working: the factorisation line is a method mark in its own right. Roots quoted straight from a calculator with no working earn nothing in 9709.

Connected rates of change

When two quantities both change with time, the chain rule links their rates:

\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}

A typical P1 question: the radius of a circle increases at 0.2 cm s⁻¹; find how fast the area grows when $r = 5$ . From $A = \pi r^2$ , $\tfrac{dA}{dr} = 2\pi r$ , so $\tfrac{dA}{dt} = 2\pi(5) \times 0.2 = 2\pi$ cm² s⁻¹. Write the chain-rule statement first — it is usually its own mark.

The mistakes that lose the marks

1.Giving the tangent when the normal was asked (and vice versa)
Cambridge examiner reports name this mix-up every session. Underline the word in the question, and after computing $m$ , write one explicit line: $m_{\text{normal}} = -1/m$ .
2.Forgetting the chain-rule factor
$(2x^2-5)^6$ does not differentiate to $6(2x^2-5)^5$ — the $\times 4x$ is where the mark lives. If the bracket’s inside isn’t just $x$ , a factor is owed.
3.Not rewriting roots and fractions before differentiating
Convert $\sqrt{x}$ to $x^{1/2}$ and $\tfrac{k}{x^n}$ to $kx^{-n}$ first. Attempting the power rule on $2/x^2$ by sight produces sign errors almost every time.
4.Solving dy/dx = 0 on the calculator with no working
9709’s standing rule: unsupported answers from a calculator earn no marks. Show the factorisation or formula line before stating roots.
5.Testing nature with y-values instead of the second derivative
Substituting the stationary $x$ into $y$ tells you the height, not the nature. Use $\tfrac{d^2y}{dx^2}$ (or a signed gradient table) and state the conclusion with its evidence.

Common questions

How do you differentiate a fraction like 3/x²?

Rewrite as a negative power first: $3x^{-2}$ , then apply the power rule: $-6x^{-3} = -\tfrac{6}{x^3}$ .

What's the difference between dy/dx and d²y/dx²?

$\tfrac{dy}{dx}$ is the gradient; $\tfrac{d^2y}{dx^2}$ is the gradient of the gradient — differentiate twice. The first finds stationary points; the second classifies them.

Is the product rule in Pure 1?

No — products, quotients, and the derivatives of $e^x$ , $\ln x$ and the trig functions all arrive in Pure 3. P1 differentiation is powers of $x$ plus the chain rule on bracketed powers.

Practise differentiation against real mark schemes

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