5. Implicit Differentiation

Video coming soon4 worked examples

Theory

When $y$ is defined implicitly (the equation is not solved for $y$ ), differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ . Each $y$ -term picks up a factor of $\frac{dy}{dx}$ by the chain rule; then collect and solve.

\frac{d}{dx}\bigl(y^n\bigr) = n\,y^{n-1}\frac{dy}{dx}

The Golden Rule: every time you differentiate a term containing $y$ , multiply by $\frac{dy}{dx}$ ; then gather all $\frac{dy}{dx}$ terms on one side and factor.

⚠️ Common Examiner Traps

Dropping $\frac{dy}{dx}$ : differentiating a $y$ -term without the chain-rule factor is the most common error.
Mixed terms: a term like $xy$ needs the product rule: $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$ .
Solving slips: factor $\frac{dy}{dx}$ carefully before dividing.

Worked examples

Example 1

Find $\dfrac{dy}{dx}$ for the circle $x^2 + y^2 = 25$ .

Step 1: Differentiate both sides with respect to $x$ :

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

Step 2: Solve for $\frac{dy}{dx}$ :

\frac{dy}{dx} = -\frac{x}{y}

Example 2

Find $\dfrac{dy}{dx}$ for $x^2 + xy + y^2 = 7$ .

Step 1: Differentiate term by term, using the product rule on $xy$ :

2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0

Step 2: Gather the $\frac{dy}{dx}$ terms:

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

Step 3: Solve:

\frac{dy}{dx} = -\frac{2x + y}{x + 2y}

Example 3

Find the equation of the tangent to the curve $x^2 + 2xy - y^2 = 7$ at the point $(2, 1)$ .

Step 1: Confirm the point lies on the curve:

(2)^2 + 2(2)(1) - (1)^2 = 4 + 4 - 1 = 7 \ \checkmark

Step 2: Differentiate implicitly, using the product rule on the $2xy$ term:

2x + 2\left(y + x\frac{dy}{dx}\right) - 2y\frac{dy}{dx} = 0

Step 3: Gather the $\frac{dy}{dx}$ terms and factorise:

\frac{dy}{dx}\bigl(2x - 2y\bigr) = -(2x + 2y) \implies \frac{dy}{dx} = -\frac{x + y}{x - y}

Step 4: Substitute $(2,1)$ to get the gradient of the tangent:

\frac{dy}{dx} = -\frac{2 + 1}{2 - 1} = -3

Step 5: Use $y - b = m(x - a)$ :

y - 1 = -3(x - 2) \implies y - 1 = -3x + 6 \implies 3x + y - 7 = 0

Example 4

Show that $(2, 8)$ is a stationary point on the curve $x^2 + y^2 - 4x - 6y = 12$ , and determine its nature.

Step 1: Check the point lies on the curve:

2^2 + 8^2 - 4(2) - 6(8) = 4 + 64 - 8 - 48 = 12 \ \checkmark

Step 2: Differentiate implicitly and make $\frac{dy}{dx}$ the subject:

2x + 2y\frac{dy}{dx} - 4 - 6\frac{dy}{dx} = 0 \implies \frac{dy}{dx}\bigl(2y - 6\bigr) = 4 - 2x

\frac{dy}{dx} = \frac{2 - x}{y - 3}

Step 3: Substitute $(2,8)$ . The numerator vanishes, so the point is stationary:

\frac{dy}{dx} = \frac{2 - 2}{8 - 3} = \frac{0}{5} = 0 \ \checkmark

Step 4: For the nature, differentiate again. Writing $p = \frac{dy}{dx}$ , the rearranged equation is $p(y - 3) = 2 - x$ , so by the product rule:

\frac{dp}{dx}\bigl(y - 3\bigr) + p \cdot p = -1

Step 5: Substitute $y = 8$ and $p = 0$ , which removes the $p^2$ term:

\frac{dp}{dx}\bigl(8 - 3\bigr) + 0 = -1 \implies 5\frac{d^2y}{dx^2} = -1 \implies \frac{d^2y}{dx^2} = -\frac{1}{5}

Step 6: Since $\frac{d^2y}{dx^2} < 0$ , the point $(2,8)$ is a maximum turning point.

Sense check: completing the square gives $(x-2)^2 + (y-3)^2 = 25$ — a circle of centre $(2,3)$ and radius $5$ . The point $(2,8)$ is the very top of that circle, so a maximum is exactly what we should expect.