Differentiation from first principles, the power rule, the chain, product and quotient rules, derivatives of standard functions including exponentials, logarithms and trigonometric functions, and implicit and parametric differentiation.

What this dot point is asking

OCR wants you to differentiate from first principles for simple polynomials, use the power rule, apply the chain, product and quotient rules, differentiate the standard functions ($e^{kx}$, $\ln x$, $\sin x$, $\cos x$, $\tan x$), and differentiate implicitly and parametrically. This is the toolkit for every later application: stationary points, tangents, rates of change, optimisation and curve sketching.

The answer

First principles

The derivative is the limit of the gradient of a chord as it shrinks to a point.

The rules and standard derivatives

The standard derivatives (with $x$ in radians) are $\dfrac{d}{dx}(e^{kx}) = ke^{kx}$, $\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$, $\dfrac{d}{dx}(\sin x) = \cos x$, $\dfrac{d}{dx}(\cos x) = -\sin x$ and $\dfrac{d}{dx}(\tan x) = \sec^2 x$.

The chain rule

The chain rule differentiates a composite "outer first, then times the inside derivative". It is the most-used rule and the most common source of dropped marks.

Implicit and parametric differentiation

For an implicit relation, differentiate every term in $x$ and treat $y$ as a function of $x$ (so $\dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx}$), then make $\dfrac{dy}{dx}$ the subject. For a parametric curve given by $x(t)$ and $y(t)$, use $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$.

Examples in context

Combining the rules

Many derivatives need two rules at once, for example a product where one factor is itself a composite. Differentiate piece by piece and keep your working laid out so method marks survive a slip.

A parametric gradient

Try this

Q1. Differentiate $y = e^{3x}\cos x$. [3 marks]

• Cue. Product rule: $\dfrac{dy}{dx} = 3e^{3x}\cos x - e^{3x}\sin x = e^{3x}(3\cos x - \sin x)$.

Q2. A curve is given by $y^2 = 4x$. Find $\dfrac{dy}{dx}$ in terms of $y$. [2 marks]

• Cue. $2y\dfrac{dy}{dx} = 4$, so $\dfrac{dy}{dx} = \dfrac{2}{y}$.