Implicit Differentiation Calculator: Find dy/dx with Steps
How to use the implicit differentiation calculator
Enter an equation relating x and y, such as x^2+y^2=25. The calculator differentiates both sides with respect to x, treating y as a function of x (so every y term picks up a dy/dx factor via the chain rule), then solves for dy/dx.
Worked example
For x^2+y^2=25, differentiating gives 2x+2y(dy/dx)=0, so dy/dx=-x/y. At the point (3,4) the slope is -3/4.
See this in action with implicit differentiation on a circle, worked step by step for x^2+y^2=25.
Frequently asked questions
When y is not isolated, for example in x^2+y^2=25 or xy=1, implicit differentiation finds dy/dx without first solving for y.
Because y depends on x, the chain rule applies: d/dx of y^2 is 2y times dy/dx, the rate at which y itself changes with x.
The method behind the answer
Differentiate every term on both sides with respect to x. Terms in x alone use the normal rules. Terms in y use the chain rule, because y is itself a function of x, so each y term gains a dy/dx factor. Mixed terms such as 3xy need the product rule, since both factors change. Then collect every dy/dx term on one side, factor dy/dx out, and divide.
Read it as: copy the power rule result for y, then multiply by dy/dx. Forgetting that final factor is the most common reason an implicit derivative comes out wrong, so check for it in every y term before you solve.
Example: Find dy/dx for x^2 + 3xy + y^2 = 5, and the slope at (1, 1)
Step 1, Differentiate each term: d/dx gives 2x + 3y + 3x dy/dx + 2y dy/dx = 0. The middle pair comes from the product rule on 3xy.
Step 2, Collect dy/dx terms: (3x + 2y) dy/dx = -(2x + 3y).
Step 3, Solve for dy/dx: Divide by (3x + 2y): dy/dx = -(2x + 3y)/(3x + 2y).
Step 4, Evaluate at the point: At (1, 1), dy/dx = -(2 + 3)/(3 + 2) = -1. The tangent slope is -1, and (1, 1) does satisfy the equation, since 1 + 3 + 1 = 5.
When implicit differentiation is the right tool
Use it for curves that are not simple graphs of y against x, such as loops and ovals, or relations where solving for y gives two branches or a messy square root. It is also the practical choice for tangent lines to such curves, and it prepares the way for related rates problems.
Common mistakes
- Missing dy/dx: Differentiating y^2 as 2y instead of 2y dy/dx drops the chain rule factor.
- Half a product rule: For xy, the derivative is y + x dy/dx, not just y.
- Dividing too early: Collect all dy/dx terms and factor first, then divide once.
- Using a point off the curve: Check the point satisfies the original equation before reporting a slope.
More questions
Yes. If the coefficient of dy/dx becomes 0 at that point while the other side is not 0, the formula divides by 0. The tangent line there is vertical, so the slope is undefined.
Find dy/dx as a formula in x and y first, then substitute the coordinates. Check first that the point lies on the curve by testing it in the original equation.
Yes. Differentiate the dy/dx expression again, apply the product and chain rules where needed, then substitute the first derivative back in to keep the answer in terms of x and y.