Implicit Differentiation Calculator

Differentiate an equation where y is not isolated, collect the y′ terms, and solve for dy/dx.

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Enter an implicit equation, such as x^2 + xy + y^2 = 7

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Differentiate the relation without isolating y

An implicit equation can describe a curve even when solving explicitly for y is awkward or splits the curve into several branches. Implicit differentiation works with the relation as written.

Because y depends on x, every derivative of an expression involving y brings in dy/dx through the chain rule. After differentiating both sides, gather those derivative terms and solve the resulting equation for the slope.

How to use the implicit differentiation calculator

Enter the problem as written

Type or paste the full expression. You can also upload a clear photo or PDF and check the extracted text before solving.

Read the working, not only the answer

Each transformation is separated and explained so you can compare it with your own method.

Ask about any step

Continue in the same solution to request another method, check a restriction, or ask why a rule applies.

Implicit derivative from F(x,y)

For a differentiable relation F(x,y) = 0, the partial derivatives give the local slope wherever Fᵧ is not zero.

A reliable way to work through it

Differentiate both sides

Treat y as y(x), using the chain rule for powers and functions of y and the product rule for terms such as xy.

Collect every y′ term

Move terms containing dy/dx to one side and factor out the common derivative.

Check the requested point

Confirm the point lies on the original curve before substituting coordinates into the slope formula.

Worked example

Find dy/dx for x² + xy + y² = 7

Differentiate both sides and use the product rule on xy.

Move the terms without y′ to the other side.

Factor out y′.

Divide by x + 2y where it is not zero.

The implicit derivative is -(2x + y)/(x + 2y). A zero denominator may indicate a vertical tangent or a singular point.

Common mistakes to check

Treating y as a constant

The equation makes y depend on x, so differentiating y contributes dy/dx.

Missing the chain-rule factor

The derivative of y² is 2y dy/dx, not only 2y.

Skipping the point check

A slope at a point is meaningful only if that point satisfies the original relation.

Questions students ask

Why does y′ appear when differentiating y?

Since y depends on x, the chain rule contributes dy/dx whenever an expression involving y is differentiated.

Could I solve for y first?

Sometimes, but that can create several branches or harder algebra. Implicit differentiation works directly with the curve.

How do I find the slope at a point?

Verify the point satisfies the equation, find dy/dx, then substitute both coordinates.

What can indicate a vertical tangent?

In -Fₓ/Fᵧ, a zero Fᵧ with nonzero Fₓ is a common vertical-tangent case. If both vanish, more analysis is needed.