Calculus · step 8 of 13

Implicit differentiation & related rates

When y is defined implicitly by an equation rather than solved for, differentiate both sides with respect to x and treat y as a function of x — so every y term picks up a dy/dx by the chain rule. Related rates apply the same idea with respect to time: differentiate a geometric relation and substitute the known rates.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Differentiate both sidesEvery term containing y contributes dy/dx by the chain rule. d/dx(y²) = 2y·dy/dx.
Use the product rule on mixed termsd/dx(xy) = y + x·dy/dx. Forgetting this term is the standard error.
Collect and solve for dy/dxGather all dy/dx terms on one side and factor.
For related rates, differentiate with respect to tWrite the geometric relation first, differentiate, then substitute the given values — substituting too early freezes a variable that is changing.

Worked example

A ladder 5 m long leans on a wall. The foot slides out at 0.3 m/s. How fast is the top falling when the foot is 3 m from the wall?

Relation: x² + y² = 25, with x the distance from the wall and y the height.
Differentiate with respect to t: 2x(dx/dt) + 2y(dy/dt) = 0.
When x = 3, y = √(25 − 9) = 4.
Substitute: 2(3)(0.3) + 2(4)(dy/dt) = 0, so 1.8 + 8(dy/dt) = 0.

Answer. dy/dt = −0.225 m/s — the top slides down at 0.225 metres per second.

Where marks get dropped

These are the specific errors that cost credit on implicit differentiation & related rates questions — QED's rubric penalises each of them separately.

Forgetting the dy/dx factor when differentiating a y term. d/dx(y²) is 2y·dy/dx, not 2y.
Substituting the specific values before differentiating, which turns a variable into a constant and gives zero rates.
Dropping the sign. A decreasing quantity has a negative rate, and the sign is part of the answer.

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Implicit differentiation & related rates — frequently asked questions

Why does y get a dy/dx?

Because y is a function of x, so differentiating any expression in y uses the chain rule. Treating y as a constant is what loses the term.

How do I set up a related rates problem?

Draw the picture, name the varying quantities, write the equation relating them, differentiate with respect to t, then substitute the instantaneous values LAST.

Can I always solve implicitly instead of explicitly?

Implicit differentiation works even when solving for y is impossible, such as x³ + y³ = 6xy. That generality is the main reason for the technique.

The rest of Calculus

Limits, derivatives, integrals and their applications. Each subtopic below has its own method, worked example and mark-losing traps.

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