Implicit Differentiation Calculator — Step-by-Step Solutions

The Implicit Differentiation Calculator finds dy/dx from equations where y is not explicitly solved for x. It handles circles, ellipses, polynomial curves, and general implicit equations.

Enter an equation with both x and y to see the derivative computed using the chain rule and partial derivatives. You can also evaluate the slope at a specific point on the curve.

Equation in x and y

Enter an equation with both x and y. Use "=" for equations, or the expression is set equal to 0. Common forms: circles (x^2 + y^2 = r^2), ellipses (x^2/a^2 + y^2/b^2 = 1), polynomials (x^2*y + y^3 = 1).

Your calculations will appear here

How to Use the Implicit Differentiation Calculator

Type your equation in x and y in the box, for example x^2 + y^2 = 25. Use the caret for powers and an explicit * between variables, so write x*y rather than xy. If you leave out the equals sign, the expression is treated as equal to 0.
Pick a mode: Find dy/dx gives the derivative along with the partial derivatives dF/dx and dF/dy, Step-by-Step Chain Rule explains how each term is differentiated, and Slope at Point evaluates dy/dx at a specific point.
For Slope at Point, enter the x and y coordinates. The point should lie on the curve. The calculator also gives the equation of the tangent line there.
The output may be arranged differently from your textbook form, for example as a single fraction. Simplify or compare by plugging in values.

Formula and Theory

When y is defined implicitly by an equation such as x^2 + y^2 = 25, you cannot always solve for y, but you can still find dy/dx. Differentiate both sides with respect to x, treating y as a function of x. Every time you differentiate a term with y, the chain rule adds a factor of dy/dx: d/dx[y^2] = 2y x dy/dx.

After differentiating, collect the dy/dx terms on one side, factor dy/dx out, and divide. The same result comes from the formula dy/dx = -F_x / F_y, where the equation is written as F(x, y) = 0 and F_x and F_y are the partial derivatives. The calculator uses this formula, and it needs F_y not to be zero. Where F_y = 0 the tangent line is vertical and dy/dx is undefined.

The product rule applies to terms such as x*y: d/dx[xy] = y + x dy/dx. The answer is usually a function of both x and y, because a single x can belong to several points on the curve; for a circle, each x has an upper and a lower point with opposite slopes.

Worked Examples

Example 1: Slope of a circle at a point

Problem: Find dy/dx for x^2 + y^2 = 25 and the slope of the tangent at (3, 4).

Solution: Differentiate both sides: 2x + 2y dy/dx = 0. Solve: dy/dx = -x / y. At (3, 4) the slope is -3/4 = -0.75. Check that (3, 4) is on the circle: 9 + 16 = 25. Tangent line: y - 4 = -0.75(x - 3), so y = -0.75x + 6.25.

Answer: dy/dx = -x/y; slope -0.75 at (3, 4)

Example 2: A curve with a product term

Problem: Find dy/dx for x^2 + xy + y^2 = 7 and the slope at (2, 1).

Solution: Differentiate: 2x + (y + x dy/dx) + 2y dy/dx = 0. Group: (x + 2y) dy/dx = -(2x + y). So dy/dx = -(2x + y) / (x + 2y). Check the point: 4 + 2 + 1 = 7. At (2, 1): -(4 + 1) / (2 + 2) = -5/4 = -1.25.

Answer: dy/dx = -(2x + y)/(x + 2y); slope -1.25 at (2, 1)

Example 3: Folium of Descartes

Problem: Find dy/dx for x^3 + y^3 = 6xy and the slope at (3, 3).

Solution: Differentiate: 3x^2 + 3y^2 dy/dx = 6y + 6x dy/dx. Move the dy/dx terms to the left: (3y^2 - 6x) dy/dx = 6y - 3x^2. So dy/dx = (6y - 3x^2) / (3y^2 - 6x) = (2y - x^2) / (y^2 - 2x). The point is on the curve: 27 + 27 = 54 = 6 x 9. At (3, 3): (6 - 9) / (9 - 6) = -3/3 = -1.

Answer: dy/dx = (2y - x^2)/(y^2 - 2x); slope -1 at (3, 3)

Example 4: With a trigonometric term

Problem: Find dy/dx for sin(y) = x and evaluate at (0, 0).

Solution: Differentiate both sides: cos(y) dy/dx = 1, so dy/dx = 1 / cos(y). At y = 0, cos(0) = 1, so the slope is 1. This is the derivative of y = arcsin(x) at x = 0.

Answer: dy/dx = 1/cos(y); slope 1 at (0, 0)

Frequently Asked Questions

What is implicit differentiation?

It is a way to find dy/dx when y is not isolated in the equation. You differentiate every term with respect to x, apply the chain rule to terms containing y, and solve for dy/dx.

When do I need implicit differentiation?

Use it for curves like circles and ellipses, or equations like x^3 + y^3 = 6xy, where y cannot be written as a simple function of x. It is also how the derivatives of inverse functions are found.

Why is there a dy/dx next to every y term?

Because y depends on x. By the chain rule, d/dx[y^n] = n y^(n-1) x dy/dx. The factor dy/dx is what you solve for at the end.

Why does my answer contain both x and y?

The slope depends on which point of the curve you are at, and a given x can match more than one y. To get a number, evaluate at a specific point (x, y) that satisfies the equation.

What if the denominator is zero?

Then the tangent is vertical at that point and dy/dx is undefined. For the circle x^2 + y^2 = 25 this happens at (5, 0) and (-5, 0), where y = 0.

How do I write multiplication of x and y?

Use an asterisk, as in x*y or 6*x*y, and use ^ for powers. Writing xy would be read as one variable called xy.