The formula this page uses
Differentiate both sides with respect to x, then solve for dy/dx. d/dx (y²) = 2y · dy/dx
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Implicit Derivative Calculator
Solve implicit derivative problems with clear steps, notation, and a final check.
d/dx
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Calculus
Implicit Derivative Calculator explained
The short version
- Implicit differentiation finds dy/dx when the equation never gets solved for y.
- Every time you differentiate a y term, the chain rule attaches a dy/dx to it.
- Then you gather every dy/dx on one side and factor it out, exactly like solving a linear equation.
What each part means
SymbolWhat it means
x — Independent variableThe variable everything is differentiated with respect to.
y — Dependent variableTreated as a hidden function of x, which is why the chain rule fires.
dy/dx — The unknownThe slope of the curve. It usually contains both x and y.
(x₀, y₀) — The pointA specific point on the curve. Substitute both coordinates to get a number.
Show your work: a full example
- The curvex² + y² = 25, a circle of radius 5, at the point (3, 4)
- Differentiate both sides with respect to x2x + 2y · dy/dx = 0
- Move the x term across2y · dy/dx = −2x
- Divide by 2ydy/dx = −x / y
- Substitute the point (3, 4)dy/dx = −3 / 4 = −0.75
- Write the tangent liney − 4 = −0.75(x − 3), so y = −0.75x + 6.25
- Sanity checkthe radius to (3, 4) has slope 4/3, and −0.75 × 4/3 = −1, so the tangent is perpendicular to it as it must be
A second, different case
- A different case: dy/dx appears on both sidesx³ + y³ = 6xy at the point (3, 3), which does satisfy it since 27 + 27 = 54 = 6(9)
- Differentiate, using the product rule on 6xy3x² + 3y² · dy/dx = 6y + 6x · dy/dx
- Collect the dy/dx terms on the left3y² · dy/dx − 6x · dy/dx = 6y − 3x²
- Factor out dy/dxdy/dx (3y² − 6x) = 6y − 3x²
- Dividedy/dx = (6y − 3x²) / (3y² − 6x) = (2y − x²) / (y² − 2x)
- Substitute (3, 3)(6 − 9) / (9 − 6) = −3 / 3
- The slope−1
Copy-ready example
x² + y² = 25, a circle of radius 5, at the point (3, 4)
The curve
Questions about the Implicit Derivative Calculator
Why can I not just solve for y first?
Sometimes you can, and for x² + y² = 25 you could write y = √(25 − x²). But x³ + y³ = 6xy has no clean solved form, and even the circle would force you to pick a branch and track a sign. Implicit differentiation avoids both problems.
Why does dy/dx usually contain both x and y?
Because an implicit equation can pass through the same x at several heights. The slope at x = 3 on the circle is −0.75 at the top and +0.75 at the bottom, so the formula needs y to know which point you mean.
What happens when the denominator is zero?
The tangent line is vertical there. On the circle, dy/dx = −x/y is undefined at (5, 0) and (−5, 0), which are precisely the two points where the circle turns straight up.
Where does implicit differentiation get used outside curve sketching?
Related rates. When a ladder slides down a wall, x² + y² stays constant, and differentiating with respect to time instead of x gives the relationship between the two speeds directly.