Implicit differentiation, explained with examples

By Ashraf Bhura, Co-Founder, STEM-Tutor · Updated October 2026

Some curves, such as circles, cannot be written as a single y = f(x). Implicit differentiation lets you find dy/dx anyway, by differentiating both sides of the equation as it stands.

Explicit and implicit equations

An explicit equation gives y directly in terms of x, such as y = x² + 1. An implicit equation mixes x and y together, such as x² + y² = 25 or x³ + y³ = 6xy. Sometimes you could rearrange for y, but often it is messy or impossible, and a circle needs two separate functions (the top half and the bottom half). Implicit differentiation avoids all of that.

The key idea

Treat y as an unknown function of x. When you differentiate any term containing y with respect to x, use the chain rule: differentiate as if y were the variable, then multiply by dy/dx. So

• d/dx (y²) = 2y · dy/dx
• d/dx (sin y) = cos y · dy/dx
• d/dx (e^y) = e^y · dy/dx
• d/dx (xy) = y + x · dy/dx, by the product rule

Terms with only x are differentiated as normal. Constants still differentiate to zero.

The four-step method

1. Differentiate every term on both sides with respect to x, adding dy/dx to each y term.
2. Move all terms containing dy/dx to one side and everything else to the other.
3. Factor out dy/dx.
4. Divide to make dy/dx the subject.

Worked examples

Example 1: a circle

Find dy/dx for x² + y² = 25, and the gradient at (3, 4).

• Differentiate: 2x + 2y(dy/dx) = 0.
• dy/dx = −x/y. At (3, 4): −3/4.
• Check: the radius to (3, 4) has gradient 4/3, and the tangent is perpendicular to it, with gradient −3/4.

Example 2: a product term

Find dy/dx for x² + xy + y² = 7.

• 2x + (y + x dy/dx) + 2y dy/dx = 0.
• dy/dx (x + 2y) = −2x − y, so dy/dx = −(2x + y)/(x + 2y).

Example 3: trig

Find dy/dx for sin y = x.

• cos y · dy/dx = 1, so dy/dx = 1/cos y.
• Since sin y = x, cos y = √(1 − x²) for −π/2 < y < π/2, giving dy/dx = 1/√(1 − x²). This is how the derivative of arcsin x is proved.

Example 4: exponentials

Find dy/dx for e^y = x² + y.

• e^y dy/dx = 2x + dy/dx.
• dy/dx (e^y − 1) = 2x, so dy/dx = 2x/(e^y − 1).

Example 5: the folium of Descartes

Find the gradient of x³ + y³ = 6xy at (3, 3).

• 3x² + 3y² dy/dx = 6y + 6x dy/dx.
• dy/dx = (6y − 3x²)/(3y² − 6x) = (2y − x²)/(y² − 2x).
• At (3, 3): (6 − 9)/(9 − 6) = −1.

Example 6: a tangent line

Find the equation of the tangent to x² + xy + y² = 7 at (1, 2).

• From Example 2: dy/dx = −(2 + 2)/(1 + 4) = −4/5.
• Tangent: y − 2 = −(4/5)(x − 1).

Example 7: horizontal and vertical tangents

For x² + y² = 25, where are the tangents horizontal and vertical?

• dy/dx = −x/y. Horizontal when the numerator is 0: x = 0, at (0, 5) and (0, −5).
• Vertical when the denominator is 0: y = 0, at (5, 0) and (−5, 0).

Example 8: a second derivative

For x² + y² = 25, find d²y/dx² in terms of y.

• dy/dx = −x/y. Quotient rule: d²y/dx² = −(y − x dy/dx)/y².
• Substitute dy/dx = −x/y: −(y + x²/y)/y² = −(y² + x²)/y³.
• Since x² + y² = 25: d²y/dx² = −25/y³. Using the original equation to simplify is a common final step.

Logarithmic differentiation

Implicit differentiation also handles functions where x appears in both the base and the power, such as y = x^x. None of the standard rules apply directly, because x^x is neither a power function nor an exponential. The trick is to take natural logs of both sides first: ln y = x ln x. Now differentiate implicitly: (1/y)(dy/dx) = ln x + 1. Multiply by y to get dy/dx = x^x(ln x + 1). The same technique simplifies derivatives of long products and quotients, because logs turn products into sums and powers into multiples.

Connection to related rates

Related rates problems use the same idea, but with time t as the variable instead of x. If a ladder's base and top satisfy x² + y² = 100, then differentiating with respect to t gives 2x(dx/dt) + 2y(dy/dt) = 0. Every variable that changes with time picks up its own rate. If you are comfortable with implicit differentiation, related rates become much easier.

How to practise

Start with simple curves like circles and xy = k until adding dy/dx feels automatic. Then move on to equations with product terms, trig and exponentials. Finally, practise full AP-style questions that ask for a tangent line, horizontal or vertical tangents, or a second derivative. Always simplify your final answer, and check it at a point on the curve if you can.

Checking your answer

When an equation can be rearranged, you can check your implicit answer explicitly. For x² + y² = 25, the top half is y = √(25 − x²), whose derivative is −x/√(25 − x²) = −x/y. It matches. Another quick check: substitute a point on the curve and see whether the gradient looks right on a sketch.

Common mistakes

• Forgetting dy/dx on y terms, for example writing d/dx(y²) = 2y.
• Missing the product rule on terms like xy or x²y.
• Differentiating a constant on the right-hand side as if it were a variable.
• Substituting the point before differentiating, which loses the variables.

Practice questions

Try each one before opening the answer.

Frequently asked questions

Is implicit differentiation on the AP Calculus exam?

Yes. It is in Unit 3 for both AB and BC, and it appears in free-response questions, often asking for a tangent line or for points where the tangent is horizontal or vertical.

Is implicit differentiation in A level Maths?

Yes, in the second year of A level Maths, usually for finding gradients, tangents and normals of curves given implicitly.

Key takeaways

• Treat y as a function of x.
• Every y term gets a dy/dx from the chain rule.
• Use the product rule on mixed terms like xy.
• Collect dy/dx terms, factor, then divide.
• Horizontal tangents: numerator zero; vertical tangents: denominator zero.

Related guides

• The chain rule
• Related rates
• A level Maths differentiation