About the Implicit Differentiation Calculator
This implicit differentiation calculator finds dy/dx for a curve defined by an equation in x and y that you cannot (or would rather not) solve for y. Type an equation such as x^2 + y^2 = 25, x^3 + y^3 = 6xy or e^(xy) = x + y and it returns the derivative as a formula, along with the partial derivatives used to build it.
Give a point on the curve and it also evaluates the slope dy/dx there, the second derivative d²y/dx² (which tells you whether the curve bends up or down), and the equations of the tangent and normal lines. It checks whether your point actually satisfies the equation so a typo in the coordinates does not go unnoticed.
It is meant for calculus students learning implicit differentiation and related rates, and for checking homework on circles, ellipses, the folium of Descartes and other implicit curves. Write products of x and y as xy or x*y, use ^ for powers and ln for natural log; trig functions use radians.
How to use the implicit differentiation calculator
- 1Type the equation using x and y, with one = sign.
- 2Enter the x and y coordinates of a point on the curve.
- 3Read dy/dx as a formula and its value at the point.
- 4Use the tangent and normal line equations, and the sign of d²y/dx² for concavity.
Formula and method
Move everything to one side so the curve is F(x, y) = 0. Differentiating both sides with respect to x and treating y as a function of x gives Fₓ + F_y·(dy/dx) = 0 by the chain rule, so dy/dx = −Fₓ/F_y. This is exactly what you get by differentiating term by term and collecting the dy/dx terms, the method taught in class.
The calculator finds the partial derivatives symbolically and simplifies the quotient. Differentiating once more gives the second-derivative formula above, evaluated at your point. Where F_y = 0 the tangent is vertical; where both partials vanish the point is singular. The tangent line is y = y₀ + m(x − x₀) and the normal has slope −1/m.
- F(x, y)
- Left side minus right side of the equation
- Fₓ, F_y
- Partial derivatives of F
- m
- Slope dy/dx at the point (x₀, y₀)
Worked examples
Circle x² + y² = 25 at (3, 4)
Differentiating gives 2x + 2y·y′ = 0, so y′ = −x/y = −3/4 at (3, 4). The second derivative is −(x² + y²)/y³ = −25/64 ≈ −0.390625 (concave down), and the tangent is y = −0.75x + 6.25.
Folium of Descartes x³ + y³ = 6xy at (3, 3)
3x² + 3y²y′ = 6y + 6xy′, so y′ = (2y − x²)/(y² − 2x) = (6 − 9)/(9 − 6) = −1 at (3, 3). The tangent is y = −x + 6 and d²y/dx² = −16/3.
Rotated ellipse x² + xy + y² = 7 at (1, 2)
2x + y + xy′ + 2yy′ = 0 gives y′ = −(2x + y)/(x + 2y) = −4/5 = −0.8 at (1, 2). The tangent line is y = −0.8x + 2.8.
Hyperbola xy = 6 at (2, 3)
By the product rule y + xy′ = 0, so y′ = −y/x = −1.5. Solving explicitly, y = 6/x gives y″ = 12/x³ = 1.5 at x = 2, matching the implicit result.
Frequently asked questions
What is implicit differentiation?+
It is a way to find dy/dx when y is not written explicitly as a function of x. You differentiate both sides of the equation with respect to x, apply the chain rule to every y term (multiplying by dy/dx), and then solve for dy/dx.
Why does dy/dx contain both x and y?+
Because the curve is defined by a relation between x and y, the slope generally depends on where you are on the curve, and that position needs both coordinates. For a circle, dy/dx = −x/y: the point (3, 4) and (3, −4) have opposite slopes.
How do you find the second derivative implicitly?+
Differentiate dy/dx again with respect to x, using the quotient rule and replacing every y′ with the first-derivative formula. For the circle x² + y² = r² this gives y″ = −r²/y³. This calculator evaluates d²y/dx² at your point automatically.
When is the tangent line vertical?+
When ∂F/∂y = 0 but ∂F/∂x ≠ 0 at the point, dy/dx would require division by zero, so the tangent is vertical — for example at (5, 0) on x² + y² = 25. If both partials are zero the point is singular.
How do I type products like xy?+
You can write xy, x*y or x y. Expressions like 3xy^2 are read as 3·x·y². Use ^ for powers, sqrt() for roots, e^(...) or exp() for exponentials and ln() for natural logs.