Implicit Differentiation

This lesson covers implicit differentiation — differentiating equations that are not written in the form y = f(x) — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to be able to find dy/dx for implicitly defined curves and use it to find tangents and normals.

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What Is Implicit Differentiation?

An explicit equation expresses y directly as a function of x:

• y = x² + 3x (explicit)

An implicit equation relates x and y without isolating y:

• x² + y² = 25 (implicit — this is a circle)
• x³ + y³ = 6xy (implicit)
• eˣ + eʸ = x + y (implicit)

For implicit equations, we differentiate both sides with respect to x, treating y as a function of x and applying the chain rule whenever we differentiate a term involving y.

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The Key Rule

When differentiating a function of y with respect to x:

d/dx[f(y)] = f'(y) × dy/dx

This is just the chain rule: y is a function of x, so differentiating f(y) with respect to x requires multiplying by dy/dx.

Term	d/dx	Explanation
y²	2y × dy/dx	Chain rule: d/dx(y²) = 2y × dy/dx
y³	3y² × dy/dx	Chain rule on y³
sin y	cos y × dy/dx	Chain rule on sin y
eʸ	eʸ × dy/dx	Chain rule on eʸ
ln y	(1/y) × dy/dx	Chain rule on ln y
x²	2x	Normal differentiation (no dy/dx needed)
xy	y + x dy/dx	Product rule (x and y are both functions of x)

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Step-by-Step Method

1. Differentiate every term with respect to x.
2. For terms involving y, apply the chain rule (multiply by dy/dx).
3. For terms involving both x and y (like xy), use the product rule.
4. Collect all dy/dx terms on one side.
5. Factor out dy/dx and solve.

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Worked Examples

Worked Example 1: Circle

Find dy/dx for x² + y² = 25.

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

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

This makes sense geometrically: for a circle centred at the origin, the tangent at (x, y) is perpendicular to the radius, which has gradient y/x. The tangent gradient -x/y is the negative reciprocal. ✓

Worked Example 2

Find dy/dx for x³ + y³ = 9.

Solution: 3x² + 3y²(dy/dx) = 0 3y²(dy/dx) = -3x² dy/dx = -x²/y²

Worked Example 3: Product Rule Needed

Find dy/dx for x²y + xy² = 6.

Solution: Differentiate term by term:

• d/dx(x²y): Use product rule with u = x² and v = y: = 2xy + x²(dy/dx)
• d/dx(xy²): Use product rule with u = x and v = y²: = y² + x × 2y(dy/dx) = y² + 2xy(dy/dx)

So: 2xy + x²(dy/dx) + y² + 2xy(dy/dx) = 0

Collect dy/dx terms: (x² + 2xy)(dy/dx) = -2xy - y²

dy/dx = -(2xy + y²) / (x² + 2xy)

This can be factorised: dy/dx = -y(2x + y) / (x(x + 2y))

Worked Example 4: With Exponentials

Find dy/dx for eˣ + eʸ = x + y.

Solution: Differentiate: eˣ + eʸ(dy/dx) = 1 + dy/dx

Collect dy/dx: eʸ(dy/dx) - dy/dx = 1 - eˣ dy/dx(eʸ - 1) = 1 - eˣ dy/dx = (1 - eˣ) / (eʸ - 1)

Exam Tip: When differentiating xy terms, you MUST use the product rule. d/dx(xy) = y + x(dy/dx), NOT just dy/dx.

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Finding Tangents and Normals to Implicit Curves

Worked Example 5

Find the equation of the tangent to the curve x² + y² = 25 at the point (3, 4).

Solution: From Example 1: dy/dx = -x/y

At (3, 4): dy/dx = -3/4

Tangent: y - 4 = (-3/4)(x - 3) 4(y - 4) = -3(x - 3) 4y - 16 = -3x + 9 3x + 4y = 25

Worked Example 6

Find the equation of the normal to x² + xy = 10 at the point (2, 3).

Solution: Differentiate: 2x + y + x(dy/dx) = 0 x(dy/dx) = -2x - y dy/dx = -(2x + y)/x

At (2, 3): dy/dx = -(4 + 3)/2 = -7/2

Normal gradient = 2/7 (negative reciprocal)

Normal: y - 3 = (2/7)(x - 2) y = (2/7)x + 17/7

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Horizontal and Vertical Tangents

Horizontal tangent (dy/dx = 0): Set the numerator of dy/dx equal to zero.

Vertical tangent (dy/dx undefined): Set the denominator of dy/dx equal to zero.

Worked Example 7

For the curve x² + y² - 4x + 2y = 0, find the points where the tangent is horizontal.

Solution: Differentiate: 2x + 2y(dy/dx) - 4 + 2(dy/dx) = 0 (2y + 2)(dy/dx) = 4 - 2x dy/dx = (4 - 2x)/(2y + 2) = (2 - x)/(y + 1)

Horizontal tangent: numerator = 0 → 2 - x = 0 → x = 2

Substitute x = 2 into the original equation: 4 + y² - 8 + 2y = 0 y² + 2y - 4 = 0 y = (-2 ± √(4 + 16))/2 = (-2 ± √20)/2 = -1 ± √5

Points: (2, -1 + √5) and (2, -1 - √5)

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Implicit Differentiation with Trig

Worked Example 8

Find dy/dx for sin(x + y) = x.

Solution: Differentiate: cos(x + y) × (1 + dy/dx) = 1

Expand: cos(x + y) + cos(x + y) × dy/dx = 1

Solve: cos(x + y) × dy/dx = 1 - cos(x + y)

dy/dx = (1 - cos(x + y)) / cos(x + y)

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Common Mistakes

Mistake	Correction
d/dx(y²) = 2y	Missing dy/dx. Correct: d/dx(y²) = 2y × dy/dx
d/dx(xy) = x × dy/dx	Missing the y term. Correct: d/dx(xy) = y + x(dy/dx) (product rule)
Differentiating the right-hand side incorrectly	If the equation is x² + y² = 25, the right side differentiates to 0 (25 is a constant)
Not collecting all dy/dx terms	Make sure every dy/dx term is on the same side before factorising

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Summary

• Implicit differentiation is used when y is not isolated as a function of x.
• Differentiate every term with respect to x, applying the chain rule to terms containing y.
• For terms like xy, use the product rule.
• Collect all dy/dx terms, factorise, and solve.
• Use the resulting dy/dx to find tangents (same gradient) and normals (negative reciprocal).
• Horizontal tangents occur where the numerator of dy/dx is zero; vertical tangents where the denominator is zero.

A-Level Deep Dive: Implicit Differentiation

Spec mapping

Edexcel 9MA0 specification, Paper 2 — Pure Mathematics, Section 9 (Differentiation), sub-strand 9.7 covers simple functions and relations defined implicitly or parametrically, for first derivative only (refer to the official specification document for exact wording). Implicit differentiation sits in Year 2 Pure content — explicitly an A2 topic — and connects forwards to section 8 (Integration, separable equations and the chain rule in reverse) and backwards to section 9.6 (the chain, product and quotient rules). Edexcel examines it on Paper 2 alongside parametric methods, and the technique is also tested synoptically on section 9.10 (related rates of change) in modelling contexts. The Edexcel formula booklet provides the chain, product and quotient rules but does not state the implicit-differentiation procedure — it must be applied as a deliberate use of the chain rule on $y$y-terms.

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Worked example with full mark scheme

Question (8 marks): A curve $C$C has equation $x^2 + xy + y^2 = 7$x2+xy+y2=7.

(a) Find $\dfrac{dy}{dx}$dxdy​ in terms of $x$x and $y$y. (5)

(b) Find an equation of the tangent to $C$C at the point $(1, 2)$(1,2), giving your answer in the form $ax + by + c = 0$ax+by+c=0 with $a, b, c$a,b,c integers. (3)

Solution with mark scheme:

(a) Step 1 — differentiate term by term with respect to $x$x.

$\dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(xy) + \dfrac{d}{dx}(y^2) = \dfrac{d}{dx}(7)$dxd​(x2)+dxd​(xy)+dxd​(y2)=dxd​(7)

M1 — recognising that every $y$y-term needs the chain rule (an extra factor of $\dfrac{dy}{dx}$dxdy​). Common error: treating $y^2$y2 as if $y$y were a constant and writing its derivative as $0$0, or differentiating to $2y$2y without the trailing $\dfrac{dy}{dx}$dxdy​.

Step 2 — apply the product rule to $xy$xy.

$\dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx}$dxd​(xy)=y+xdxdy​

M1 — correct product-rule expansion. The $y$y comes from differentiating $x$x (giving $1$1) times $y$y; the $x\dfrac{dy}{dx}$xdxdy​ comes from $x$x times the chain-rule derivative of $y$y.

Step 3 — assemble.

$2x + y + x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0$2x+y+xdxdy​+2ydxdy​=0

A1 — fully correct differentiated equation.

Step 4 — solve for $\dfrac{dy}{dx}$dxdy​.

Group $\dfrac{dy}{dx}$dxdy​ terms: $\dfrac{dy}{dx}(x + 2y) = -(2x + y)$dxdy​(x+2y)=−(2x+y).

M1 — collecting $\dfrac{dy}{dx}$dxdy​ on one side.

$\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}$dxdy​=−x+2y2x+y​

A1 — final expression.

(b) Step 1 — substitute $(1, 2)$(1,2).

$\dfrac{dy}{dx}\bigg|_{(1,2)} = -\dfrac{2(1) + 2}{1 + 2(2)} = -\dfrac{4}{5}$dxdy​​(1,2)​=−1+2(2)2(1)+2​=−54​

M1 — substitution into the gradient formula.

Step 2 — point-gradient form. $y - 2 = -\tfrac{4}{5}(x - 1)$y−2=−54​(x−1), so $5(y - 2) = -4(x - 1)$5(y−2)=−4(x−1), giving $4x + 5y - 14 = 0$4x+5y−14=0.

M1 A1 — rearranged into integer form, verified by checking the point lies on it.

Total: 8 marks (M5 A3).

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): The curve $C$C has equation $e^{2x} + xy^2 = 4y + 5$e2x+xy2=4y+5.

(a) Show that $\dfrac{dy}{dx} = \dfrac{2e^{2x} + y^2}{4 - 2xy}$dxdy​=4−2xy2e2x+y2​. (4)

(b) Find the gradient of $C$C at the point where $x = 0$x=0 and $y = -1$y=−1. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — differentiating $e^{2x}$e2x correctly to $2e^{2x}$2e2x via the chain rule.
• M1 (AO1.1b) — applying product rule to $xy^2$xy2: $y^2 + 2xy\dfrac{dy}{dx}$y2+2xydxdy​.
• M1 (AO1.1b) — differentiating RHS to $4\dfrac{dy}{dx}$4dxdy​ and assembling $2e^{2x} + y^2 + 2xy\dfrac{dy}{dx} = 4\dfrac{dy}{dx}$2e2x+y2+2xydxdy​=4dxdy​.
• A1 (AO2.1) — rearranging to the printed form, with the explicit collection of $\dfrac{dy}{dx}$dxdy​ shown.

(b)

• M1 (AO1.1b) — substituting $x = 0$x=0, $y = -1$y=−1 into the formula.
• A1 (AO1.1b) — gradient $= \dfrac{2(1) + 1}{4 - 0} = \dfrac{3}{4}$=4−02(1)+1​=43​.

Total: 6 marks split AO1 = 5, AO2 = 1. Implicit-differentiation questions are AO1-heavy: marks reward procedural correctness (chain + product rules) with a single AO2 mark for the algebraic rearrangement to a printed form.

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Synoptic links