Implicit Differentiation — Pure Mathematics 3 (9709) — Mathematics

Introduction

Most functions encountered so far are explicit — written in the form $y = f(x)$ , where $y$ is isolated on one side. However, many curves are defined by an equation linking $x$ and $y$ that cannot easily (or at all) be rearranged into explicit form. For example, the curve $x^2 + y^2 = xy + 7$ relates $x$ and $y$ but it is not straightforward to write $y$ as a function of $x$ .

Implicit differentiation is the technique that allows us to find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ directly from such an equation, without rearranging. It is a reliable source of marks in 9709 Paper 3, frequently appearing in tangent/normal questions worth 6–9 marks.

Core Concept

The key idea is to differentiate both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ and applying the chain rule whenever a term in $y$ is differentiated.

The Chain Rule for terms in $y$

For any function of $y$ :

\frac{\mathrm{d}}{\mathrm{d}x}\bigl[f(y)\bigr] = f'(y)\,\frac{\mathrm{d}y}{\mathrm{d}x}

This is because $y$ depends on $x$ , so the chain rule gives $\dfrac{\mathrm{d}}{\mathrm{d}x}[f(y)] = \dfrac{\mathrm{d}f}{\mathrm{d}y} \cdot \dfrac{\mathrm{d}y}{\mathrm{d}x}$ .

Mixed terms — the Product Rule

When a term contains both $x$ and $y$ (e.g. $xy$ , $x^2 y$ ), differentiate using the product rule:

\frac{\mathrm{d}}{\mathrm{d}x}(xy) = y + x\,\frac{\mathrm{d}y}{\mathrm{d}x}

General procedure

Differentiate every term on both sides with respect to $x$ .
Collect all terms containing $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ on one side.
Factorise and solve for $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ .
Substitute the coordinates of a given point to find the gradient at that point.

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