Mathematics 9709 · Pure Mathematics 2

Implicit Differentiation

Some curves cannot be written as $y=\mathrm{f}(x)$ . Here $x$ and $y$ are tangled together, like $x^3+5xy+y^3=16$ .
You can still differentiate. There is one new idea, and it is just the chain rule again.

$y$ is a function of $x$ you can't see, so $y$ counts as an “inside.” So every time you differentiate a $y$ -term, the chain rule adds a $\dfrac{dy}{dx}$ on the end: $\dfrac{d}{dx}(y^3)=3y^2\dfrac{dy}{dx}$ .

The fixed 4-step recipe

Differentiate both sides term by term with respect to $x$ . For a $y$ -term use the chain rule (the $\frac{dy}{dx}$ tag); for a mixed $xy$ -term use the product rule.
Collect every $\frac{dy}{dx}$ term on one side, everything else on the other.
Factor out $\frac{dy}{dx}$ .
Divide to get $\frac{dy}{dx}$ in terms of $x$ and $y$ .

The answer really does contain both $x$ and $y$ . That is normal for implicit curves, not a mistake. (Worked Example 4.12)

Worked example

9709/32 Jun 2023 Q7: show $\dfrac{dy}{dx} = -\dfrac{3x+2y}{2x+3y}$ for $3x^2 + 4xy + 3y^2 = 5$

Differentiate each term. $3x^2\to 6x$ . The $4xy$ needs the product rule: $4xy\to 4x\frac{dy}{dx}+4y$ . The $3y^2$ needs the chain rule: $3y^2\to 6y\frac{dy}{dx}$ . RHS: $5\to 0$ .
So $6x+4x\dfrac{dy}{dx}+4y+6y\dfrac{dy}{dx}=0$ .
Collect the $\frac{dy}{dx}$ terms and factor: $(4x+6y)\dfrac{dy}{dx}=-(6x+4y)$ .
Divide, then cancel the shared factor of $2$ : $\dfrac{dy}{dx}=\dfrac{-(6x+4y)}{4x+6y}=-\dfrac{3x+2y}{2x+3y}$ .

Answer

\dfrac{dy}{dx}=-\dfrac{3x+2y}{2x+3y}

This is a “show that”, so the answer is printed. That means every mark is in the tagged steps: the $\frac{dy}{dx}$ on the $y$ -terms and the two product-rule pieces on $4xy$ . Never skip them. (9709/32 May/June 2023 Q7)

Worked example

Real question: $x^2y - ay^2 = 4a^3$ , show $\dfrac{dy}{dx} = \dfrac{2xy}{2ay - x^2}$

9709/31 M/J 2023 Q5(a). The constant $a$ is just a number, so treat it like one, and watch the two mixed $xy$ -terms.

$x^2y$ is a product: $\dfrac{d}{dx}(x^2y)=x^2\dfrac{dy}{dx}+2xy$ . The $ay^2$ uses the chain rule: $\dfrac{d}{dx}(ay^2)=2ay\dfrac{dy}{dx}$ . RHS $4a^3$ is constant, so $\to 0$ .
So $x^2\dfrac{dy}{dx}+2xy-2ay\dfrac{dy}{dx}=0$ .
Collect and factor: $(x^2-2ay)\dfrac{dy}{dx}=-2xy$ .
Divide, then flip both signs to match the printed form: $\dfrac{dy}{dx}=\dfrac{-2xy}{x^2-2ay}=\dfrac{2xy}{2ay-x^2}$ .

Answer

\dfrac{dy}{dx}=\dfrac{2xy}{2ay-x^2}

Multiplying top and bottom by $-1$ matches the printed answer, and examiners expect it. (9709/31 May/June 2023 Q5)

Common mistake

The three things that go wrong most with implicit:

Forgetting the $\frac{dy}{dx}$ tag on a $y$ -term — writing $\frac{d}{dx}(y^3)=3y^2$ instead of $3y^2\frac{dy}{dx}$ . This is THE implicit mistake.
Treating a mixed term as one piece. $5xy$ is a product, so $\frac{d}{dx}(5xy)=5x\frac{dy}{dx}+5y$ — two terms, not one.
Algebra slips when collecting and factoring $\frac{dy}{dx}$ .

Using dy/dx: gradients, tangents, stationary points

For a gradient at a point you need both coordinates: plug in $x$ and $y$ (usually given).
For a tangent or normal, find the gradient there, then write the usual line.
For a horizontal tangent, set the top of $\frac{dy}{dx}$ to zero; for a vertical tangent, set the bottom to zero, then solve it together with the curve.

Worked example

Real question: $2y^2 + 3xy + x = x^2$ , show no horizontal tangent exists

9709/32 F/M 2024 Q6. Part (a) gives $\dfrac{dy}{dx}=\dfrac{2x-3y-1}{4y+3x}$ ; part (b) asks why the curve is never flat.

A horizontal tangent needs the top to vanish: $2x-3y-1=0$ .
Solve that together with the curve. Sub $x=\tfrac{3y+1}{2}$ into $2y^2+3xy+x=x^2$ ; it reduces to a quadratic in $y$ whose discriminant is negative.
No real $y$ satisfies both, so the gradient is never zero: the curve has no tangent parallel to the $x$ -axis.

Answer

\text{no real solution} \Rightarrow \text{no horizontal tangent}

“Show it never happens” means “show the equations have no real solution”. A negative discriminant is the cleanest proof. (9709/32 Feb/March 2024 Q6)

Your turn— tap to reveal the worked answer (9709/31 May/June 2022 Q8)

$x^3+y^3+2xy+8=0$ ; express $\frac{dy}{dx}$ in terms of $x$ and $y$ . Differentiate (the $2xy$ is a product): $3x^2+3y^2\frac{dy}{dx}+2x\frac{dy}{dx}+2y=0$ .

Answer

\dfrac{dy}{dx}=-\dfrac{3x^2+2y}{3y^2+2x}

Implicit also mixes with logs and exponentials, such as $\ln(x+y)=3x^2y$ (9709-style) and $ye^{2x}-y^2e^{x}=2$ (9709-style). But the steps never change.

PreviousDifferentiating sin, cos and tan NextParametric Differentiation

Implicit Differentiation

Some curves cannot be written as $y=\mathrm{f}(x)$ . Here $x$ and $y$ are tangled together, like $x^3+5xy+y^3=16$ .
You can still differentiate. There is one new idea, and it is just the chain rule again.

The fixed 4-step recipe

Differentiate both sides term by term with respect to $x$ . For a $y$ -term use the chain rule (the $\frac{dy}{dx}$ tag); for a mixed $xy$ -term use the product rule.
Collect every $\frac{dy}{dx}$ term on one side, everything else on the other.
Factor out $\frac{dy}{dx}$ .
Divide to get $\frac{dy}{dx}$ in terms of $x$ and $y$ .

The answer really does contain both $x$ and $y$ . That is normal for implicit curves, not a mistake. (Worked Example 4.12)

Worked example

9709/32 Jun 2023 Q7: show $\dfrac{dy}{dx} = -\dfrac{3x+2y}{2x+3y}$ for $3x^2 + 4xy + 3y^2 = 5$

Differentiate each term. $3x^2\to 6x$ . The $4xy$ needs the product rule: $4xy\to 4x\frac{dy}{dx}+4y$ . The $3y^2$ needs the chain rule: $3y^2\to 6y\frac{dy}{dx}$ . RHS: $5\to 0$ .
So $6x+4x\dfrac{dy}{dx}+4y+6y\dfrac{dy}{dx}=0$ .
Collect the $\frac{dy}{dx}$ terms and factor: $(4x+6y)\dfrac{dy}{dx}=-(6x+4y)$ .
Divide, then cancel the shared factor of $2$ : $\dfrac{dy}{dx}=\dfrac{-(6x+4y)}{4x+6y}=-\dfrac{3x+2y}{2x+3y}$ .

Answer

\dfrac{dy}{dx}=-\dfrac{3x+2y}{2x+3y}

Worked example

Real question: $x^2y - ay^2 = 4a^3$ , show $\dfrac{dy}{dx} = \dfrac{2xy}{2ay - x^2}$

9709/31 M/J 2023 Q5(a). The constant $a$ is just a number, so treat it like one, and watch the two mixed $xy$ -terms.

$x^2y$ is a product: $\dfrac{d}{dx}(x^2y)=x^2\dfrac{dy}{dx}+2xy$ . The $ay^2$ uses the chain rule: $\dfrac{d}{dx}(ay^2)=2ay\dfrac{dy}{dx}$ . RHS $4a^3$ is constant, so $\to 0$ .
So $x^2\dfrac{dy}{dx}+2xy-2ay\dfrac{dy}{dx}=0$ .
Collect and factor: $(x^2-2ay)\dfrac{dy}{dx}=-2xy$ .
Divide, then flip both signs to match the printed form: $\dfrac{dy}{dx}=\dfrac{-2xy}{x^2-2ay}=\dfrac{2xy}{2ay-x^2}$ .

Answer

\dfrac{dy}{dx}=\dfrac{2xy}{2ay-x^2}

Multiplying top and bottom by $-1$ matches the printed answer, and examiners expect it. (9709/31 May/June 2023 Q5)

Common mistake

The three things that go wrong most with implicit:

Forgetting the $\frac{dy}{dx}$ tag on a $y$ -term — writing $\frac{d}{dx}(y^3)=3y^2$ instead of $3y^2\frac{dy}{dx}$ . This is THE implicit mistake.
Treating a mixed term as one piece. $5xy$ is a product, so $\frac{d}{dx}(5xy)=5x\frac{dy}{dx}+5y$ — two terms, not one.
Algebra slips when collecting and factoring $\frac{dy}{dx}$ .

Using dy/dx: gradients, tangents, stationary points

For a gradient at a point you need both coordinates: plug in $x$ and $y$ (usually given).
For a tangent or normal, find the gradient there, then write the usual line.
For a horizontal tangent, set the top of $\frac{dy}{dx}$ to zero; for a vertical tangent, set the bottom to zero, then solve it together with the curve.

Worked example

Real question: $2y^2 + 3xy + x = x^2$ , show no horizontal tangent exists

9709/32 F/M 2024 Q6. Part (a) gives $\dfrac{dy}{dx}=\dfrac{2x-3y-1}{4y+3x}$ ; part (b) asks why the curve is never flat.

A horizontal tangent needs the top to vanish: $2x-3y-1=0$ .
Solve that together with the curve. Sub $x=\tfrac{3y+1}{2}$ into $2y^2+3xy+x=x^2$ ; it reduces to a quadratic in $y$ whose discriminant is negative.
No real $y$ satisfies both, so the gradient is never zero: the curve has no tangent parallel to the $x$ -axis.

Answer

\text{no real solution} \Rightarrow \text{no horizontal tangent}

“Show it never happens” means “show the equations have no real solution”. A negative discriminant is the cleanest proof. (9709/32 Feb/March 2024 Q6)

Your turn— tap to reveal the worked answer (9709/31 May/June 2022 Q8)

$x^3+y^3+2xy+8=0$ ; express $\frac{dy}{dx}$ in terms of $x$ and $y$ . Differentiate (the $2xy$ is a product): $3x^2+3y^2\frac{dy}{dx}+2x\frac{dy}{dx}+2y=0$ .

Answer

\dfrac{dy}{dx}=-\dfrac{3x^2+2y}{3y^2+2x}

Implicit also mixes with logs and exponentials, such as $\ln(x+y)=3x^2y$ (9709-style) and $ye^{2x}-y^2e^{x}=2$ (9709-style). But the steps never change.