Implicit Differentiation

Builds on: aqa-alevel-maths-pure-2 / differentiation-techniques — this lesson is the A2-level deep dive.

This lesson covers implicit differentiation, a technique for finding dy/dx when the equation relating x and y is not (or cannot easily be) written in the form y = f(x). Many important curves — circles, ellipses, and other higher-degree curves — are most naturally expressed implicitly, and this technique is essential for A-Level Mathematics.

Spec Mapping — AQA 7357 Section G (Differentiation). This lesson covers the implicit-differentiation content of Section G, at the depth required for A2-level synoptic questions. Refer to the official AQA specification document for exact wording.

What Is an Implicit Equation?

An explicit equation expresses y directly in terms of x: for example, y = x² + 3x.

An implicit equation relates x and y without solving for y: for example, x² + y² = 25 or x³ + y³ = 6xy.

For implicit equations, we cannot always isolate y, so we differentiate both sides of the equation with respect to x, treating y as a function of x.

The Key Rule

When differentiating a term involving y with respect to x, use the chain rule:

d/dx [f(y)] = f'(y) \times dy/dx

In particular:

d/dx(y²) = 2y × dy/dx
d/dx(y³) = 3y² × dy/dx
d/dx(yⁿ) = nyⁿ⁻¹ × dy/dx
d/dx(sin y) = cos y × dy/dx
d/dx(eʸ) = eʸ × dy/dx

Basic Examples

Example 1: Find dy/dx given x² + y² = 25.

\begin{aligned} &\text{Differentiate both sides with respect to } x: \\ &2x + 2y\frac{dy}{dx} = 0 \\[6pt] &\text{Solve for } \frac{dy}{dx}: \\ &2y\frac{dy}{dx} = -2x \\ &\frac{dy}{dx} = -\frac{x}{y} \end{aligned}

Interpretation: For the circle x² + y² = 25, at the point (3, 4):

dy/dx = -3/4

This gives the gradient of the tangent at (3, 4), which is perpendicular to the radius (gradient 4/3), as expected.

Example 2: Find dy/dx given x³ + y³ = 9.

\begin{aligned} 3x^{2} + 3y^{2}\frac{dy}{dx} &= 0 \\ \frac{dy}{dx} &= -\frac{x^{2}}{y^{2}} \end{aligned}

Example 3: Find dy/dx given 2x² + 3y² = 14.

\begin{aligned} 4x + 6y\frac{dy}{dx} &= 0 \\ \frac{dy}{dx} &= -\frac{4x}{6y} = -\frac{2x}{3y} \end{aligned}

Product Terms (xy Terms)

When the equation contains a term like xy, you must use the product rule:

\frac{d}{dx}(xy) = x\frac{dy}{dx} + y \times 1 = x\frac{dy}{dx} + y

More generally:

\begin{aligned} \frac{d}{dx}(x^{2}y) &= x^{2}\frac{dy}{dx} + 2xy \\[4pt] \frac{d}{dx}(xy^{2}) &= x \times 2y\frac{dy}{dx} + y^{2} = 2xy\frac{dy}{dx} + y^{2} \end{aligned}

Example 4: Find dy/dx given x² + xy + y² = 7.

\begin{aligned} &\text{Differentiate term by term:} \\ \frac{d}{dx}(x^{2}) &= 2x \\ \frac{d}{dx}(xy) &= x\frac{dy}{dx} + y &&[\text{product rule}] \\ \frac{d}{dx}(y^{2}) &= 2y\frac{dy}{dx} &&[\text{chain rule}] \\ \frac{d}{dx}(7) &= 0 \\[6pt] \text{So: } \ &2x + x\frac{dy}{dx} + y + 2y\frac{dy}{dx} = 0 \\[6pt] &\text{Collect } \frac{dy}{dx} \text{ terms:} \\ &(x + 2y)\frac{dy}{dx} = -2x - y \\[4pt] &\frac{dy}{dx} = -\frac{2x + y}{x + 2y} \end{aligned}

Example 5: Find dy/dx given x²y + xy² = 6.

\begin{aligned} \frac{d}{dx}(x^{2}y) &= x^{2}\frac{dy}{dx} + 2xy &&[\text{product rule}] \\ \frac{d}{dx}(xy^{2}) &= x \times 2y\frac{dy}{dx} + y^{2} &&[\text{product rule + chain rule}] \\[6pt] x^{2}\frac{dy}{dx} + 2xy + 2xy\frac{dy}{dx} + y^{2} &= 0 \\[4pt] (x^{2} + 2xy)\frac{dy}{dx} &= -2xy - y^{2} \\[4pt] \frac{dy}{dx} &= -\frac{2xy + y^{2}}{x^{2} + 2xy} \\[4pt] \frac{dy}{dx} &= -\frac{y(2x + y)}{x(x + 2y)} \end{aligned}

Finding the Gradient at a Specific Point

Once you have an expression for dy/dx, substitute the coordinates of the point.

Example 6: Given x² + y² − 6x + 4y = 12, find the gradient at the point (7, 1).

\begin{aligned} \text{Differentiate: } \ &2x + 2y\frac{dy}{dx} - 6 + 4\frac{dy}{dx} = 0 \\ &(2y + 4)\frac{dy}{dx} = 6 - 2x \\ &\frac{dy}{dx} = \frac{6 - 2x}{2y + 4} \\[6pt] &\text{At } (7, 1): \frac{dy}{dx} = \frac{6 - 14}{2 + 4} = -\frac{8}{6} = -\frac{4}{3} \end{aligned}

First check that (7, 1) lies on the curve: 49 + 1 − 42 + 4 = 12. ✓

Tangent and Normal Lines

Example 7: Find the equation of the tangent to the curve x² + 2xy − y² = 8 at the point (2, 2).

\begin{aligned} &\text{Check: } 4 + 8 - 4 = 8. \;\checkmark \\[6pt] \text{Differentiate: } \ &2x + 2\left[x\frac{dy}{dx} + y\right] - 2y\frac{dy}{dx} = 0 \\ &2x + 2x\frac{dy}{dx} + 2y - 2y\frac{dy}{dx} = 0 \\ &(2x - 2y)\frac{dy}{dx} = -2x - 2y \\ &\frac{dy}{dx} = -\frac{2x + 2y}{2x - 2y} \\ &\frac{dy}{dx} = -\frac{x + y}{x - y} \\[6pt] &\text{At } (2, 2): \frac{dy}{dx} = -\frac{4}{0} \to \text{undefined!} \end{aligned}

The gradient is undefined, which means the tangent is vertical: x = 2.

Example 8: Find the equation of the tangent to y² = x³ at the point (1, 1).

\begin{aligned} \text{Check} &: 1 = 1 \;\checkmark \\ \text{Differentiate} &: 2y(dy/dx) = 3x^{2} \\ dy/dx &= 3x^{2} / (2y) \\ \text{At } (1, 1) &: dy/dx = 3/2 \\ \text{Tangent} &: y - 1 = (3/2)(x - 1) \\ &\;\;\; 2y - 2 = 3x - 3 \\ &\;\;\; 3x - 2y - 1 = 0 \end{aligned}

Stationary Points

Stationary points occur where dy/dx = 0. From the implicit derivative, this usually means the numerator equals zero (while the denominator is non-zero).

Example 9: Find the stationary points of the curve x² + y² + 6x − 2y = 6.

\begin{aligned} \text{Differentiate: } \ &2x + 2y\frac{dy}{dx} + 6 - 2\frac{dy}{dx} = 0 \\ &(2y - 2)\frac{dy}{dx} = -2x - 6 \\ &\frac{dy}{dx} = -\frac{2x + 6}{2y - 2} = -\frac{x + 3}{y - 1} \\[6pt] &\text{Stationary when numerator} = 0: x + 3 = 0 \implies x = -3 \\[6pt] &\text{Substitute } x = -3 \text{ into the original equation:} \\ &9 + y^{2} - 18 - 2y = 6 \\ &y^{2} - 2y - 15 = 0 \\ &(y - 5)(y + 3) = 0 \\ &y = 5 \ \text{or}\ y = -3 \\[6pt] &\text{Stationary points: } (-3,\ 5) \text{ and } (-3,\ -3). \end{aligned}

Check denominators: at (−3, 5), y − 1 = 4 ≠ 0. ✓ At (−3, −3), y − 1 = −4 ≠ 0. ✓

(This curve is actually the circle (x + 3)² + (y − 1)² = 16, and the stationary points are at the top and bottom of the circle.)

Implicit Differentiation with Exponentials and Logarithms

Example 10: Find dy/dx given eˣʸ = x + y.

\begin{aligned} \text{Differentiate: } \ &e^{xy} \times \frac{d}{dx}(xy) = 1 + \frac{dy}{dx} \\ &e^{xy} \times \left(x\frac{dy}{dx} + y\right) = 1 + \frac{dy}{dx} \\ &xe^{xy}\frac{dy}{dx} + ye^{xy} = 1 + \frac{dy}{dx} \\ &xe^{xy}\frac{dy}{dx} - \frac{dy}{dx} = 1 - ye^{xy} \\ &(xe^{xy} - 1)\frac{dy}{dx} = 1 - ye^{xy} \\ &\frac{dy}{dx} = \frac{1 - ye^{xy}}{xe^{xy} - 1} \end{aligned}

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked Examination-Style Problem

Problem: The curve C has equation x² − 4xy + y² + 27 = 0.

(a) Show that dy/dx = (4y − 2x) / (2y − 4x).

(b) Find the coordinates of the points on C where the tangent is parallel to the x-axis.

Solution:

(a)

\begin{aligned} \text{Differentiate: } \ &2x - 4\left[x\frac{dy}{dx} + y\right] + 2y\frac{dy}{dx} = 0 \\ &2x - 4x\frac{dy}{dx} - 4y + 2y\frac{dy}{dx} = 0 \\ &(2y - 4x)\frac{dy}{dx} = 4y - 2x \\ &\frac{dy}{dx} = \frac{4y - 2x}{2y - 4x} \;\checkmark \end{aligned}

(b) Tangent parallel to x-axis means dy/dx = 0:

\begin{aligned} &4y - 2x = 0 \implies y = \frac{x}{2} \\[6pt] &\text{Substitute } y = \tfrac{x}{2} \text{ into } x^{2} - 4xy + y^{2} + 27 = 0: \\ &x^{2} - 4x\left(\tfrac{x}{2}\right) + \left(\tfrac{x}{2}\right)^{2} + 27 = 0 \\ &x^{2} - 2x^{2} + \frac{x^{2}}{4} + 27 = 0 \\ &-x^{2} + \frac{x^{2}}{4} + 27 = 0 \\ &-\frac{3x^{2}}{4} + 27 = 0 \\ &x^{2} = 36 \\ &x = \pm 6 \\[6pt] &\text{When } x = 6: y = 3. \quad \text{When } x = -6: y = -3. \\[4pt] &\text{Points: } (6,\ 3) \text{ and } (-6,\ -3). \end{aligned}

Synoptic Links

This topic connects to:

Differentiation techniques (aqa-alevel-maths-pure-2 / differentiation-techniques) — chain, product and quotient rules applied here term-by-term to y as a function of x.
Parametric differentiation (aqa-alevel-maths-coordinate-geometry-depth / parametric-differentiation) — a sister technique for finding dy/dx without solving for y explicitly.
Rates of change (aqa-alevel-maths-calculus-applications / rates-of-change) — implicit differentiation underlies related-rates problems where x and y change with time.

Summary

Implicit differentiation: differentiate every term with respect to x, applying the chain rule to y terms (multiply by dy/dx).
Product rule is needed for xy-type terms: d/dx(xy) = x(dy/dx) + y.
After differentiating, collect all dy/dx terms on one side and solve.
Stationary points: set dy/dx = 0 (numerator = 0, denominator ≠ 0).
Vertical tangents: denominator = 0 (numerator ≠ 0).
Always verify that your point lies on the curve before finding the gradient there.

Exam Tip: When implicitly differentiating, work term by term and write each derivative clearly. A common mistake is forgetting the dy/dx factor when differentiating y terms. For product terms like xy, write out the product rule explicitly. Always simplify your final expression for dy/dx and state the gradient at the specific point asked for.

A-Level Deep Dive: Implicit Differentiation

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section G: Differentiation (Year 2) covers simple functions and relations defined implicitly … for first derivative only (refer to the official specification document for exact wording). Implicit differentiation is the Year 2 extension of the chain rule that turns the AS-level rule $\frac{d}{dx}x^n = nx^{n-1}$ into the toolkit needed for any curve specified as $F(x, y) = 0$ rather than $y = f(x)$ . It is examined within Paper 2 alongside parametric differentiation (also Section G), and its results feed directly into Section H (Integration, where rearranging $\frac{dy}{dx}$ for separable equations relies on implicit reasoning), Section F (Coordinate geometry — circles, ellipses, lemniscates as implicit curves) and Section J (Numerical methods, where Newton–Raphson on $F(x, y) = 0$ uses partial derivatives in disguise). The AQA formula booklet provides standard derivatives but no implicit-differentiation template — the chain-rule application must be reproduced from scratch on every question.

Worked example with full mark scheme

Question (8 marks): The curve $C$ has equation $x^2 + xy + y^2 = 7$ .

(a) Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ . (4)

(b) Find an equation of the tangent to $C$ at the point $P(1, 2)$ , giving your answer in the form $ax + by + c = 0$ where $a$ , $b$ , $c$ are integers. (4)

Solution with mark scheme:

(a) Step 1 — differentiate term-by-term, treating $y$ as a function of $x$ .

For $x^2$ : standard derivative $2x$ .

For $xy$ : this is a product of two functions of $x$ (since $y = y(x)$ ). Apply the product rule: $\frac{d}{dx}(xy) = 1 \cdot y + x \cdot \frac{dy}{dx} = y + x\frac{dy}{dx}$ .

For $y^2$ : apply the chain rule. $\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$ .

For the constant $7$ : derivative is $0$ .

M1 — attempting to differentiate every term, with the chain-rule factor $\frac{dy}{dx}$ appearing on at least one $y$ -term. Candidates who write $\frac{d}{dx}(y^2) = 2y$ and stop have missed the point of the topic and lose this M1.

M1 — correct product rule on $xy$ , producing both $y$ and $x\frac{dy}{dx}$ .

Step 2 — assemble the equation.

$2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$

A1 — fully correct differentiated equation, all four terms in place with correct signs.

Step 3 — solve for $\frac{dy}{dx}$ .

Group the $\frac{dy}{dx}$ terms:

$(x + 2y)\frac{dy}{dx} = -(2x + y)$ $\frac{dy}{dx} = -\dfrac{2x + y}{x + 2y}$

A1 — correctly rearranged with $\frac{dy}{dx}$ subject. Equivalent forms (e.g. $\dfrac{-(2x + y)}{x + 2y}$ or $\dfrac{2x + y}{-(x + 2y)}$ ) are accepted.

(b) Step 1 — verify $P(1, 2)$ lies on $C$ .

$1^2 + 1 \cdot 2 + 2^2 = 1 + 2 + 4 = 7$ . Yes. B1 — most mark schemes award an implicit B1 here for any working that confirms the point.

Step 2 — evaluate $\frac{dy}{dx}$ at $P$ .

$\left.\frac{dy}{dx}\right|_{(1,2)} = -\dfrac{2(1) + 2}{1 + 2(2)} = -\dfrac{4}{5}$

M1 — correct substitution into the result of part (a).

Step 3 — form the tangent equation.

Using point-slope form with gradient $-\tfrac{4}{5}$ and point $(1, 2)$ :

$y - 2 = -\dfrac{4}{5}(x - 1)$ $5(y - 2) = -4(x - 1)$ $5y - 10 = -4x + 4$ $4x + 5y - 14 = 0$

M1 — correct application of point-slope form with the evaluated gradient.

A1 — answer $4x + 5y - 14 = 0$ in the requested integer form.

Total: 8 marks (M4 A3 B1).

Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): A curve has equation $\ln(xy) + y = 2x$ for $x, y > 0$ .

(a) Show that $\dfrac{dy}{dx} = \dfrac{y(2x - 1)}{x(y + 1)}$ . (4)

(b) Find the equation of the normal to the curve at the point where $x = 1$ and $y = 1$ . (2)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — splitting $\ln(xy) = \ln x + \ln y$ before differentiating, or differentiating $\ln(xy)$ directly using the chain rule as $\dfrac{1}{xy} \cdot \frac{d}{dx}(xy)$ .
M1 (AO1.1b) — applying the chain rule to $\ln y$ : $\frac{d}{dx}(\ln y) = \dfrac{1}{y}\frac{dy}{dx}$ . Differentiating $y$ on the right gives $\frac{dy}{dx}$ .
A1 (AO1.1b) — fully correct differentiated equation $\dfrac{1}{x} + \dfrac{1}{y}\frac{dy}{dx} + \frac{dy}{dx} = 2$ .
A1 (AO2.1) — clean rearrangement to the printed form, including multiplication through by $xy$ and grouping $(y + 1)\frac{dy}{dx}$ .

(b)

M1 (AO1.1b) — substituting $x = 1$ , $y = 1$ into part (a)'s gradient and inverting-and-negating for the normal: tangent gradient $\dfrac{1 \cdot 1}{1 \cdot 2} = \tfrac{1}{2}$ , normal gradient $-2$ .
A1 (AO1.1b) — normal equation $y - 1 = -2(x - 1)$ , i.e. $y = -2x + 3$ or $2x + y - 3 = 0$ .

Total: 6 marks split AO1 = 5, AO2 = 1. AQA implicit-differentiation questions are AO1-dominated — the technique itself is procedural — with AO2 marks reserved for clean rearrangement and the conceptual jump between tangent and normal gradients.

Synoptic links

Connects to:

Section G — Chain rule: implicit differentiation is the chain rule applied to compositions $y(x)$ that you cannot solve explicitly. The factor $\frac{dy}{dx}$ on every $y$ -term is the chain-rule "inner derivative" $\frac{dy}{dx}$ from $\frac{d}{dx}f(y) = f'(y)\frac{dy}{dx}$ .
Section G — Product rule: any term containing both $x$ and $y$ (such as $xy$ , $x^2 y$ , $xe^y$ ) requires the product rule alongside the chain rule. Implicit differentiation is the natural stress-test of product-and-chain combination.
Section F — Circles and conics: the circle $x^2 + y^2 = r^2$ has implicit gradient $\frac{dy}{dx} = -\frac{x}{y}$ , which is the negative reciprocal of the radius gradient $\frac{y}{x}$ — the geometric statement "tangent perpendicular to radius" falls out of implicit differentiation in two lines.
Section G — Tangent and normal lines: every tangent/normal question at A-Level uses the gradient $m_T$ then $m_N = -1/m_T$ . Implicit differentiation extends this from $y = f(x)$ curves to any $F(x, y) = 0$ curve, including those (such as the lemniscate $(x^2 + y^2)^2 = a^2(x^2 - y^2)$ ) that cannot be written as $y = f(x)$ .
Section G — Logarithmic differentiation: to differentiate $y = x^x$ or $y = x^{\sin x}$ , take logs first ( $\ln y = x \ln x$ ) and then differentiate implicitly. This technique reduces hard expressions to standard chain-rule applications and is the canonical A* extension of the topic.

Implicit Differentiation

Implicit Differentiation

What Is an Implicit Equation?

The Key Rule

Basic Examples

Product Terms (xy Terms)

Finding the Gradient at a Specific Point

Tangent and Normal Lines

Stationary Points

Implicit Differentiation with Exponentials and Logarithms

Specimen Question (Modelled on the AQA Paper Format)

Worked Examination-Style Problem

Synoptic Links

Summary

A-Level Deep Dive: Implicit Differentiation

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 2 format

Synoptic links

Mark-scheme literacy

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