Derivatives
GradesPre-Calculus & CalculusDerivativesImplicit Differentiation

Implicit Differentiation

📉 Derivatives
🔄

Implicit Differentiation

Differentiate equations where y is defined implicitly as a function of x.

When an equation defines y implicitly (not solved for y), we differentiate both sides with respect to x. Whenever y appears, we apply the chain rule: dy/dx appears automatically because y is a function of x.

🔑

Key: treat y as a function of x. Every time you differentiate a y-term, multiply by dy/dx (chain rule). Then solve algebraically for dy/dx.

Implicit Differentiation Steps

1Differentiate both sides of the equation with respect to x.
2For each term with y, apply the chain rule: d/dx[yⁿ] = n·yⁿ⁻¹·(dy/dx).
3Group all terms with dy/dx on one side.
4Factor out dy/dx and divide to isolate it.
✏️Differentiate x² + y² = 25
Differentiate: 2x + 2y·(dy/dx) = 0. Solve: dy/dx = −2x/(2y) = −x/y. At the point (3, 4): slope = −3/4. This is the slope of the tangent to the circle at (3, 4).
x2+y2=25  ⟹  2x+2ydydx=0  ⟹  dydx=−xyx^2 + y^2 = 25 \implies 2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}
✏️Product of x and y
Differentiate x²y + y³ = 6. Product rule on x²y: 2xy + x²(dy/dx). Chain rule on y³: 3y²(dy/dx). Total: 2xy + x²(dy/dx) + 3y²(dy/dx) = 0. Factor: dy/dx(x² + 3y²) = −2xy. So dy/dx = −2xy/(x² + 3y²).
✏️ Try It!

For the curve x³ + y³ = 9, what is dy/dx?

Practice with CalcVerse
d/dxDerivative Solver
Take Quiz 📝 — 24 Questions