Differentiation from first principles, the rules for powers, the chain, product and quotient rules, derivatives of standard functions, stationary points and their nature, and connected rates of change.

What this dot point is asking

AQA wants you to differentiate from first principles, use the power, chain, product and quotient rules, differentiate standard functions including trigonometric, exponential and logarithmic ones, find and classify stationary points, and apply differentiation to tangents, normals, connected rates of change and optimisation. Optimisation questions, where you build a function, differentiate, and justify a maximum or minimum, are among the most heavily weighted in Paper 1.

First principles

The derivative is defined as the limit $f'(x) = \displaystyle\lim_{h \to 0}\dfrac{f(x + h) - f(x)}{h}$. This is the gradient of the chord between two nearby points as they merge. Applying it to $f(x) = x^2$: $\dfrac{(x + h)^2 - x^2}{h} = \dfrac{2xh + h^2}{h} = 2x + h$, which tends to $2x$ as $h \to 0$, so $f'(x) = 2x$. AQA may ask for a first-principles derivation, so know the limit definition and how to take the limit cleanly.

The rules

Standard derivatives to memorise include $\frac{d}{dx}(e^x) = e^x$, $\frac{d}{dx}(e^{kx}) = ke^{kx}$, $\frac{d}{dx}(\ln x) = \frac{1}{x}$, $\frac{d}{dx}(\sin x) = \cos x$ and $\frac{d}{dx}(\cos x) = -\sin x$ (with $x$ in radians). Combining these with the chain rule covers most exam derivatives, for example $\frac{d}{dx}\sin(3x) = 3\cos(3x)$.

Tangents and normals

The gradient of the curve at a point is the value of $\frac{dy}{dx}$ there. The tangent at $(x_1, y_1)$ has that gradient, and the normal is perpendicular, with gradient the negative reciprocal. These link directly to the coordinate geometry topic.

Stationary points and their nature

Connected rates and optimisation

If two quantities both depend on time, the chain rule links their rates: $\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt}$. This is how you find, say, the rate a balloon's volume grows given the rate its radius grows. In optimisation you write the quantity to be maximised or minimised as a function of one variable (using a constraint to eliminate any others), differentiate, set the derivative to zero, solve, and confirm the nature of the stationary point with the second derivative.

Increasing, decreasing and convexity

The sign of the first derivative describes the behaviour of the curve: where $\frac{dy}{dx} > 0$ the function is increasing, where $\frac{dy}{dx} < 0$ it is decreasing, and stationary points separate these regions. Questions often ask you to find the set of values of $x$ for which a function is increasing, which means solving the inequality $\frac{dy}{dx} > 0$. The second derivative describes the bending: where $\frac{d^2 y}{dx^2} > 0$ the curve is convex (concave up) and where $\frac{d^2 y}{dx^2} < 0$ it is concave (concave down). A point where the concavity changes sign is a point of inflection.

A reliable optimisation method

Optimisation questions reward a clear structure. First, identify the quantity to optimise and write it as a formula. Second, use the given constraint to eliminate any extra variables so the quantity depends on one variable only (this is the step the question's earlier "show that" part usually sets up). Third, differentiate and set the derivative to zero to find the candidate value. Fourth, justify that it is the required maximum or minimum, normally with the second-derivative test, and finally compute the optimal value of the original quantity. Skipping the justification, or forgetting to convert back from the variable to the quantity asked for, are the commonest ways to lose the final marks.