Unit 3: Differentiation — Composite, Implicit & Inverse Functions

Implicit Differentiation

Differentiating equations that are not solved explicitly for y.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

The Chain Rule

Differentiating Without Solving for y First

The equation x² + y² = 25 defines a circle — but it isn't solved for y, and solving it would produce two separate branches, y = √(25 − x²) and y = −√(25 − x²). Before reading on: is there a way to find the slope of the tangent line to this circle without ever splitting it into two functions?

Definition — Implicit Differentiation

Implicit differentiation treats y as an unspecified function of x and differentiates both sides of an equation with respect to x directly, applying the chain rule to every y-term (since y itself depends on x, differentiating a y-term produces a factor of dy/dx). Solving the resulting equation for dy/dx gives the slope, without ever isolating y explicitly.

Worked Example — Differentiating a Circle Implicitly

Differentiate x² + y² = 25 with respect to x. The left side: d/dx[x²] + d/dx[y²] = 2x + 2y(dy/dx), using the chain rule on y² (y is a function of x, so its derivative carries a dy/dx factor). The right side: d/dx[25] = 0. So 2x + 2y(dy/dx) = 0, giving dy/dx = −x/y. At the point (3, 4) — check: 3² + 4² = 25, confirmed on the circle — dy/dx = −3/4. This matches geometric intuition: the radius from the origin to (3, 4) has slope 4/3, and a tangent line to a circle is perpendicular to its radius, so its slope should be the negative reciprocal, −3/4 — exactly what implicit differentiation found.

x^2 + y^2 = 25 \;\Rightarrow\; \frac{dy}{dx} = -\frac{x}{y}

Worked Example — An Implicit Equation Needing the Product Rule Too

Find dy/dx for x²y + y³ = 10 at the point (1, 2) — check: 1²(2) + 2³ = 2 + 8 = 10, confirmed on the curve. Differentiate term by term: d/dx[x²y] needs the product rule (x² and y are both functions of x): 2xy + x²(dy/dx). d/dx[y³] needs the chain rule: 3y²(dy/dx). The right side, d/dx[10] = 0. So 2xy + x²(dy/dx) + 3y²(dy/dx) = 0. Collect the dy/dx terms: dy/dx · (x² + 3y²) = −2xy, so dy/dx = −2xy/(x² + 3y²). At (1, 2): dy/dx = −2(1)(2)/(1² + 3(2²)) = −4/(1 + 12) = −4/13.

Tip

Every term containing y needs the chain rule applied — differentiate as usual, then multiply by dy/dx, since y is standing in for an unspecified function of x. Terms with only x differentiate normally, with no extra factor.

Common Mistakes

Differentiating y² as 2y and forgetting the chain rule's extra dy/dx factor.
y is a function of x, not the independent variable itself — d/dx[y²] = 2y · (dy/dx), the same chain rule already used for d/dx[(3x+1)⁵], just with y playing the role of the inner function.
Forgetting the product rule on a mixed term like x²y, and differentiating it as though only one factor depended on x.
Both x² and y depend on x, so a term like x²y needs the full product rule: d/dx[x²y] = 2xy + x²(dy/dx), not just one of those two pieces.

Key Takeaways

Implicit differentiation applies the chain rule to every y-term when differentiating an equation with respect to x, producing a dy/dx factor each time.
Solving the differentiated equation for dy/dx gives the slope without ever isolating y — useful when a relation can't be cleanly solved for y, or splits into multiple branches.
A mixed term like x²y needs the product rule in addition to the chain rule, since both factors depend on x.

Summary

Implicit differentiation handles equations, not just functions. The next lesson uses the same chain-rule idea to differentiate a function's inverse — including the inverse trigonometric functions.

Track your progress

← The Chain Rule Back to Unit 3: Differentiation — Composite, Implicit & Inverse Functions Derivatives of Inverse Functions→