Implicit Differentiation

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.

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) × 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.

Differentiate both sides with respect to x:
2x + 2y(dy/dx) = 0

Solve for dy/dx:
2y(dy/dx) = −2x
dy/dx = −x/y

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.

3x² + 3y²(dy/dx) = 0
dy/dx = −x²/y²

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

4x + 6y(dy/dx) = 0
dy/dx = −4x/(6y) = −2x/(3y)

Product Terms (xy Terms)

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

d/dx(xy) = x(dy/dx) + y × 1 = x(dy/dx) + y

More generally:

d/dx(x²y) = x²(dy/dx) + 2xy
d/dx(xy²) = x × 2y(dy/dx) + y² = 2xy(dy/dx) + y²

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

Differentiate term by term:
d/dx(x²) = 2x
d/dx(xy) = x(dy/dx) + y          [product rule]
d/dx(y²) = 2y(dy/dx)             [chain rule]
d/dx(7) = 0

So: 2x + x(dy/dx) + y + 2y(dy/dx) = 0

Collect dy/dx terms:
(x + 2y)(dy/dx) = −2x − y

dy/dx = −(2x + y) / (x + 2y)

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

d/dx(x²y) = x²(dy/dx) + 2xy     [product rule]
d/dx(xy²) = x × 2y(dy/dx) + y²  [product rule + chain rule]

x²(dy/dx) + 2xy + 2xy(dy/dx) + y² = 0

(x² + 2xy)(dy/dx) = −2xy − y²

dy/dx = −(2xy + y²) / (x² + 2xy)
dy/dx = −y(2x + y) / (x(x + 2y))

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).

Differentiate: 2x + 2y(dy/dx) − 6 + 4(dy/dx) = 0
(2y + 4)(dy/dx) = 6 − 2x
dy/dx = (6 − 2x) / (2y + 4)

At (7, 1): dy/dx = (6 − 14) / (2 + 4) = −8/6 = −4/3

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).

Check: 4 + 8 − 4 = 8. ✓

Differentiate: 2x + 2[x(dy/dx) + y] − 2y(dy/dx) = 0
2x + 2x(dy/dx) + 2y − 2y(dy/dx) = 0
(2x − 2y)(dy/dx) = −2x − 2y
dy/dx = −(2x + 2y) / (2x − 2y)
dy/dx = −(x + y) / (x − y)

At (2, 2): dy/dx = −(4) / (0)  →  undefined!

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).

Check: 1 = 1. ✓

Differentiate: 2y(dy/dx) = 3x²
dy/dx = 3x² / (2y)

At (1, 1): dy/dx = 3/2

Tangent: y − 1 = (3/2)(x − 1)
         2y − 2 = 3x − 3
         3x − 2y − 1 = 0

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.

Differentiate: 2x + 2y(dy/dx) + 6 − 2(dy/dx) = 0
(2y − 2)(dy/dx) = −2x − 6
dy/dx = −(2x + 6) / (2y − 2) = −(x + 3) / (y − 1)

Stationary when numerator = 0: x + 3 = 0 ⟹ x = −3

Substitute x = −3 into the original equation:
9 + y² − 18 − 2y = 6
y² − 2y − 15 = 0
(y − 5)(y + 3) = 0
y = 5 or y = −3

Stationary points: (−3, 5) and (−3, −3).

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.

Differentiate: eˣʸ × d/dx(xy) = 1 + dy/dx
eˣʸ × (x(dy/dx) + y) = 1 + dy/dx
xeˣʸ(dy/dx) + yeˣʸ = 1 + dy/dx
xeˣʸ(dy/dx) − dy/dx = 1 − yeˣʸ
(xeˣʸ − 1)(dy/dx) = 1 − yeˣʸ
dy/dx = (1 − yeˣʸ) / (xeˣʸ − 1)

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)

Differentiate: 2x − 4[x(dy/dx) + y] + 2y(dy/dx) = 0
2x − 4x(dy/dx) − 4y + 2y(dy/dx) = 0
(2y − 4x)(dy/dx) = 4y − 2x
dy/dx = (4y − 2x) / (2y − 4x)  ✓

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

4y − 2x = 0 ⟹ y = x/2

Substitute y = x/2 into x² − 4xy + y² + 27 = 0:
x² − 4x(x/2) + (x/2)² + 27 = 0
x² − 2x² + x²/4 + 27 = 0
−x² + x²/4 + 27 = 0
−3x²/4 + 27 = 0
x² = 36
x = ±6

When x = 6: y = 3. When x = −6: y = −3.

Points: (6, 3) and (−6, −3).

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.

Mark-scheme literacy

Implicit-differentiation questions on 7357 split AO marks heavily toward AO1 in part (a) and shift toward AO1.1b/AO2 in part (b):

AO	Typical share	Earned by
AO1 (knowledge / procedure)	65–80%	Differentiating each term, applying chain rule on $y$ -terms with $\frac{dy}{dx}$ factor, applying product rule on mixed $xy$ -terms, rearranging for $\frac{dy}{dx}$
AO2 (reasoning / interpretation)	15–25%	Identifying tangent vs normal, recognising domain restrictions where $\frac{dy}{dx}$ is undefined (vertical tangents), connecting $\frac{dy}{dx} = 0$ to stationary points on the implicit curve
AO3 (problem-solving)	0–15%	Multi-step problems: find points where the tangent is parallel to a given line; locate stationary points on $F(x, y) = 0$

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

Worked Examination-Style Problem

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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