Implicit Differentiation Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Implicit Differentiation.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

A technique for finding \frac{dy}{dx} when y is defined implicitly by an equation F(x, y) = 0 rather than explicitly as y = f(x). Differentiate both sides with respect to x, treating y as a function of x, then solve for \frac{dy}{dx}.

Sometimes you can't (or don't want to) solve for y explicitly. Instead, differentiate the whole equation as-is. Every time you differentiate a y-term, attach \frac{dy}{dx} by the chain rule (since y secretly depends on x), then solve for \frac{dy}{dx}.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: The chain rule is the key: \frac{d}{dx}[y^2] = 2y \cdot \frac{dy}{dx} because y is a function of x. After differentiating, collect all \frac{dy}{dx} terms on one side and solve.

Common stuck point: Don't forget \frac{dy}{dx} every time you differentiate a term involving y. For example, \frac{d}{dx}[xy] = x\frac{dy}{dx} + y by the product rule, with \frac{dy}{dx} appearing because y depends on x.

Sense of Study hint: After differentiating, circle every dy/dx term, move them all to one side, factor out dy/dx, and divide.

Worked Examples

Example 1

easy
Find \frac{dy}{dx} for the circle x^2 + y^2 = 25 and evaluate it at the point (3, 4).

Solution

  1. 1
    Differentiate both sides with respect to x: 2x + 2y\frac{dy}{dx} = 0.
  2. 2
    Solve for \frac{dy}{dx}: \frac{dy}{dx} = -\frac{x}{y}.
  3. 3
    At (3, 4): \frac{dy}{dx} = -\frac{3}{4}.

Answer

\dfrac{dy}{dx} = -\dfrac{x}{y}; at (3, 4): slope = -\dfrac{3}{4}
Whenever a y-term is differentiated, attach \frac{dy}{dx} by the chain rule. Then collect all \frac{dy}{dx} terms and solve. The tangent to a circle at (3,4) has slope -3/4.

Example 2

hard
Find \frac{dy}{dx} for x^3 + y^3 = 6xy (folium of Descartes).

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
Find \frac{dy}{dx} for x^2 + 3y^2 = 7.

Example 2

medium
Find the equation of the tangent line to x^2y + y^3 = 2 at the point (1, 1).
← Back to Implicit Differentiation Formula Explained Practice Mode

Background Knowledge

These ideas may be useful before you work through the harder examples.

derivativechain ruledifferentiation rules