Math & Science

Implicit Derivative Calculator 2026

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dy/dx step-by-step
Polynomial implicit equations
How to use: Enter each term of the equation f(x, y) = [RHS] in the form coeff*x^a*y^b. For example, x² + y² = 25 uses two terms: (1, x^2, y^0) and (1, x^0, y^2) with RHS = 25.
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Implicit Derivative Calculator: How the Method Works

For an equation f(x, y) = c, differentiating both sides with respect to x gives: df/dx + df/dy * (dy/dx) = 0. Solving for dy/dx yields the formula below. This works for any differentiable implicit equation.

dy/dx = -(df/dx) / (df/dy)
For term: c x x^a x y^b
  df/dx = c x a x x^(a-1) x y^b
  df/dy = c x b x x^a x y^(b-1)

The formula dy/dx = -(df/dx) / (df/dy) directly uses partial derivatives of f with respect to each variable. This is the same operation computed by the Partial Derivative Calculator, which finds df/dx and df/dy separately for any polynomial in x and y.

Second Derivative Implicit Differentiation: Finding d²y/dx²

To find the second derivative d²y/dx² of an implicitly defined function, differentiate dy/dx with respect to x. Since dy/dx = -(df/dx) / (df/dy), apply the quotient rule to this expression, then substitute your earlier expression for dy/dx.

Example: x^2 + y^2 = 25
Step 1: dy/dx = -x/y (from first derivative)
Step 2: d^2y/dx^2 = d/dx[-x/y]
= -(y - x times dy/dx) / y^2 (quotient rule)
= -(y - x times (-x/y)) / y^2
= -(y + x^2/y) / y^2
= -(y^2 + x^2) / y^3
= -25/y^3  (using x^2 + y^2 = 25)

The key step is substituting the expression for dy/dx back into the result after applying the quotient rule. This eliminates dy/dx from the second derivative expression and leaves d²y/dx² in terms of x and y only. For the reverse operation of differentiation, see the Antiderivative Calculator.

Implicit Differentiation and Partial Derivatives

Implicit differentiation and partial derivatives are deeply connected. The formula dy/dx = -(df/dx) / (df/dy) can be read as: the ratio of the partial derivative of f with respect to x, divided by the partial derivative of f with respect to y, negated.

OperationWhat it computesNotationKey rule applied
Partial df/dxRate of change w.r.t. x, y held constantdf/dx or f_xPower rule, y treated as constant
Partial df/dyRate of change w.r.t. y, x held constantdf/dy or f_yPower rule, x treated as constant
Implicit dy/dxHow y changes as x changes on the curvedy/dxChain rule, dy/dx factor on y terms
Relationshipdy/dx = -(df/dx) / (df/dy)Implicit Function TheoremBoth operations used together

The Implicit Function Theorem guarantees that as long as df/dy is not zero at a point, the implicit equation F(x, y) = c locally defines y as a differentiable function of x near that point. For polynomial estimation between known data points (a related numerical technique), see the Interpolation Calculator.

Example Calculation

Equation: x² + y² = 25. Find dy/dx at (3, 4).

Terms: (1, 2, 0) = x^2   and   (1, 0, 2) = y^2, RHS = 25
df/dx = 2x    df/dy = 2y
Differentiate both sides w.r.t. x:
  2x + 2y x dy/dx = 0
Isolate dy/dx:
  dy/dx = -2x / 2y = -x/y
At (3, 4): df/dx = 6, df/dy = 8
dy/dx = -6/8 = -0.75

The slope of the tangent to the circle x² + y² = 25 at point (3, 4) is -0.75. This is consistent with the geometric fact that the tangent is perpendicular to the radius from the origin to (3, 4), which has slope 4/3.

Frequently Asked Questions

Implicit differentiation is a technique for finding dy/dx when y is not explicitly defined as a function of x. In an explicit function like y = x^2 + 3, you differentiate directly. In an implicit equation like x^2 + y^2 = 25, y is defined implicitly. You differentiate both sides with respect to x, applying the chain rule wherever y appears (each y term picks up a dy/dx factor), then solve for dy/dx.

More Math & Science Calculators

Common Examples
x^2 + y^2 = 25
Terms: (1,2,0) + (1,0,2)
dy/dx = -x/y
x^2 + xy + y^2 = 7
Terms: (1,2,0) + (1,1,1) + (1,0,2)
dy/dx = -(2x+y)/(x+2y)
x^3 + y^3 = 8
Terms: (1,3,0) + (1,0,3)
dy/dx = -x^2/y^2
Key Rule
When differentiating a term like y^n with respect to x, apply the chain rule: d/dx[y^n] = n times y^(n-1) times dy/dx. The dy/dx factor always appears when differentiating any function of y with respect to x.
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