Differentiation practice questions (A-Level Maths)

Sample questions

The one free A-Level Maths question on this topic, from the free A-Level Maths Practice Paper. Each answer is behind its own toggle; the whole paper, its PDF and the full answer key are on the exam page.

1. Question 17 marks · Differentiation & stationary points
   A curve has equation $y = 2x^3 - 9x^2 + 12x - 3$.
   $y = 2x^3 - 9x^2 + 12x - 3$
   (a)

   Find the coordinates of the stationary points of the curve.

   [4]
   (b)

   Determine the nature of each stationary point.

   [2]
   (c)

   State the range of values of $x$ for which $y$ is increasing.

   [1]
   Show answerFrom the free A-Level Maths Practice Paper →

Key terms

The definitions and formulas A-Level Maths expects for differentiation, typeset as they appear on the paper.

Differentiation from first principles

The derivative as the limit of the gradient of a chord. A first-principles question expects this definition written down, the algebra expanded, and the limit taken explicitly.

$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

Standard derivatives

The results assumed throughout; those for $\sin$, $\cos$ and $e^x$ are in the formulae booklet, the power rule and $\ln x$ are not.

$$\frac{d}{dx}x^n = nx^{n-1}, \quad \frac{d}{dx}\sin kx = k\cos kx, \quad \frac{d}{dx}\cos kx = -k\sin kx, \quad \frac{d}{dx}\ln x = \frac{1}{x}$$

Stationary points and their nature

Where $f'(x) = 0$. The second derivative classifies them: positive is a minimum, negative a maximum; if it is zero, check the sign of $f'$ on either side.

$$f'(x) = 0, \qquad f''(x) > 0 \Rightarrow \text{minimum}, \quad f''(x) < 0 \Rightarrow \text{maximum}$$

Tangents and normals

The tangent at $x = a$ has gradient $f'(a)$ and the normal has gradient $-\frac{1}{f'(a)}$; both pass through $(a, f(a))$.

$$y - f(a) = f'(a)(x - a)$$

Chain, product and quotient rules

For composite functions, products and quotients respectively. Quote the rule you use; the method mark depends on it.

$$\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}, \qquad (uv)' = u'v + uv', \qquad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$

Implicit differentiation and connected rates

Differentiate each term with respect to $x$, attaching $\frac{dy}{dx}$ to terms in $y$. For related rates, link the rates with the chain rule and substitute values after differentiating.

$$\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt}$$

Convex, concave and points of inflection

A curve is convex where $f''(x) > 0$ and concave where $f''(x) < 0$; a point of inflection is where $f''$ changes sign (zero alone is not enough).

Frequently asked questions

Other A-Level Maths topics

• Proof, algebra and functions
• Coordinate geometry
• Sequences and series
• Trigonometry
• Exponentials and logarithms
• Integration
• Numerical methods and vectors
• Sampling and data presentation
• Probability and distributions
• Hypothesis testing
• Kinematics
• Forces, Newton's laws and moments

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