Unit 3: Differentiation — Composite, Implicit & Inverse Functions

Implicit Differentiation

Differentiating implicitly defined relations and inverse functions.

Advanced25 min lesson2 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

The Chain Rule

Differentiating Without Solving for y — Including Inverses

Definition — Implicit Differentiation

Treating y as an unspecified function of x, differentiate both sides of an equation with respect to x — every y-term picks up a dy/dx factor by the chain rule. Solving for dy/dx gives the slope without ever isolating y.

Worked Example — Differentiating a Circle Implicitly

Differentiate x² + y² = 20 at the point (4, 2) — check: 16 + 4 = 20, confirmed on the circle. 2x + 2y(dy/dx) = 0, so dy/dx = −x/y. At (4, 2): dy/dx = −4/2 = −2. Check geometrically: the radius from the origin to (4, 2) has slope 2/4 = 1/2, and a circle's tangent is perpendicular to its radius, so the tangent slope should be the negative reciprocal, −2 — matching exactly.

The same idea applies to a function's inverse. Differentiating f(f⁻¹(x)) = x with the chain rule gives f'(f⁻¹(x)) · (f⁻¹)'(x) = 1, so (f⁻¹)'(x) = 1/f'(f⁻¹(x)) — the reciprocal of f' evaluated at f⁻¹(x), not at x itself.

(f^{-1})'(x) = \frac{1}{f'\big(f^{-1}(x)\big)}

Worked Example — An Inverse Derivative Without Solving for the Inverse

Let f(x) = x³ + 4. Find (f⁻¹)'(12). Solve x³ + 4 = 12: x³ = 8, x = 2, so f⁻¹(12) = 2. f'(x) = 3x², f'(2) = 12. (f⁻¹)'(12) = 1/12. Check with the explicit inverse: f⁻¹(x) = (x−4)^(1/3), (f⁻¹)'(x) = (1/3)(x−4)^(−2/3), and at x = 12: (1/3)(8)^(−2/3) = (1/3)(1/4) = 1/12 — matching.

The same implicit approach derives the inverse trig derivatives: for y = arctan(x), tan(y) = x, so sec²(y)(dy/dx) = 1, and since sec²y = 1 + tan²y = 1 + x², dy/dx = 1/(1 + x²). The analogous approach on sin(y) = x gives arcsin's derivative, 1/√(1 − x²).

Tip

In (f⁻¹)'(x) = 1/f'(f⁻¹(x)), f' is evaluated at f⁻¹(x), not at x — find f⁻¹(x) first before evaluating f'.

Common Mistakes

Forgetting the chain rule's dy/dx factor when differentiating a y-term, like treating d/dx[y²] as just 2y.
y depends on x, so d/dx[y²] = 2y · (dy/dx) — the same chain rule already used for composite functions of x alone.

Key Takeaways

Implicit differentiation applies the chain rule to every y-term, producing a dy/dx factor each time.
(f⁻¹)'(x) = 1/f'(f⁻¹(x)) follows from differentiating f(f⁻¹(x)) = x.
arctan's and arcsin's derivatives both come from implicitly differentiating their defining equations.

Summary

This closes Unit 3. Unit 4 puts differentiation to work in applied contexts, starting with quantities changing together over time.

Track your progress

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