Implicit Differentiation Calculator
Compute dy/dx for equations involving both x and y using implicit differentiation. Use this when y cannot be easily isolated as an explicit function of x.
About this calculator
Implicit differentiation finds dy/dx for an equation like F(x, y) = 0 without solving for y explicitly. For an equation with an x term, a y term, and an xy cross term — such as A·x + B·y + C·xy = k — differentiating both sides with respect to x using the chain rule on y-terms and the product rule on the xy term gives: A + B(dy/dx) + C(y + x·dy/dx) = 0. Solving for dy/dx yields: dy/dx = -(A + C·B) / (B + C·A), which matches the formula -(xCoefficient + xyCoefficient × yCoefficient) / (yCoefficient + xyCoefficient × xCoefficient). This approach generalizes to any implicitly defined curve and is foundational to related-rates problems and finding tangent lines to implicit curves.
How to use
Consider the equation 3x + 4y + 2xy = 10, so xCoefficient = 3, yCoefficient = 4, xyCoefficient = 2. Apply the formula: dy/dx = -(3 + 2 × 4) / (4 + 2 × 3) = -(3 + 8) / (4 + 6) = -11 / 10 = -1.1. The slope of the implicit curve at a general point described by these coefficients is dy/dx = -1.1. A negative value means y decreases as x increases along the curve at that location.
Frequently asked questions
What is implicit differentiation and when do you need to use it?
Implicit differentiation is a technique for computing dy/dx when x and y appear together in an equation that cannot easily be rearranged to the form y = f(x). Classic examples include circles (x² + y² = r²), ellipses, and more complex algebraic curves. Instead of isolating y — which may require choosing one of several branches — you differentiate both sides with respect to x, treating y as a function of x and applying the chain rule wherever y appears.
How does the product rule apply when differentiating an xy term implicitly?
The product rule states d/dx[u·v] = u'v + uv'. For the term C·xy, treat C as a constant, u = x, and v = y. Then d/dx[C·xy] = C·(1·y + x·dy/dx) = C·y + C·x·(dy/dx). The extra C·x·(dy/dx) term is why implicit differentiation of cross terms produces dy/dx on both sides of the equation, which you then collect and solve for. Forgetting the product rule on xy terms is the most common error in implicit differentiation.
Why can implicit differentiation give a slope that depends on both x and y coordinates?
Unlike explicit functions where dy/dx depends only on x, an implicit curve can pass through the same x value at multiple y values (think of a circle). The slope at each of those points is different, so the derivative must encode both coordinates to distinguish them. This is actually useful: it means the formula gives the exact tangent slope at a specific point (x₀, y₀) on the curve, not just a generic x-based formula, making it straightforward to find tangent and normal lines to complex curves.