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Daily Math Minute

Unit 3: Differentiation — Composite, Implicit & Inverse Functions

Derivatives of Inverse Functions

Finding derivatives of inverse functions and inverse trigonometric functions.

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

Prerequisites

  • Implicit Differentiation

Differentiating a Function's Undo-Button

f and f⁻¹ satisfy f(f⁻¹(x)) = x for every x in the appropriate domain — one always exactly undoes the other. Before reading on: if that equation holds for every x, what happens if both sides are differentiated with respect to x?

Differentiating f(f⁻¹(x)) = x, the left side needs the chain rule (f⁻¹(x) is the inner function): f'(f⁻¹(x)) · (f⁻¹)'(x) = 1, since the right side's derivative is 1. Solving for (f⁻¹)'(x) gives the inverse function derivative rule.

Definition — Derivative of an Inverse Function

(f⁻¹)'(x) = 1 / f'(f⁻¹(x)) — the derivative of the inverse at x is the reciprocal of the original function's derivative, evaluated at f⁻¹(x), not at x itself.
(f−1)′(x)=1f′(f−1(x))(f^{-1})'(x) = \frac{1}{f'\big(f^{-1}(x)\big)}

Worked Example — Finding an Inverse Derivative Without Solving for the Inverse

Let f(x) = x³ + 2. Find (f⁻¹)'(3). First find f⁻¹(3): solve x³ + 2 = 3, giving x³ = 1, so x = 1 — meaning f⁻¹(3) = 1. Then f'(x) = 3x², so f'(1) = 3. By the rule: (f⁻¹)'(3) = 1/f'(f⁻¹(3)) = 1/f'(1) = 1/3. Check by finding the inverse explicitly: f⁻¹(x) = (x − 2)^(1/3), so (f⁻¹)'(x) = (1/3)(x − 2)^(−2/3), and at x = 3: (1/3)(1)^(−2/3) = 1/3 — matching exactly.

Worked Example — Deriving the Derivative of arctan

Let y = arctan(x), so tan(y) = x by definition. Differentiate both sides implicitly with respect to x: sec²(y) · (dy/dx) = 1, so dy/dx = 1/sec²(y). Using the identity sec²y = 1 + tan²y, and tan(y) = x by the original definition, sec²(y) = 1 + x². So dy/dx = 1/(1 + x²) — the derivative of arctan, derived directly from implicit differentiation rather than stated.

The same technique — implicit differentiation of the defining equation — gives arcsin's derivative: for y = arcsin(x), sin(y) = x, so cos(y)(dy/dx) = 1, and since cos(y) = √(1 − sin²y) = √(1 − x²) (positive, since arcsin's range keeps y in [−π/2, π/2] where cosine is non-negative), dy/dx = 1/√(1 − x²).

Function Explorer

Transform: g(x) = a·f(b(x − h)) + k
1
1
0
0
Composition
Analysis (of the transformed function, in view)
y-intercept
(0, 0)
x-intercepts
(-9.42, 0), (-6.28, 0), (-3.14, 0), (0, 0), (3.14, 0), (6.28, 0), (9.42, 0)
Extrema
local min at (-7.85, -1); local max at (-4.71, 1); local min at (-1.57, -1); local max at (1.57, 1); local min at (4.71, -1); local max at (7.85, 1)
Inflection points
(-9.42, 0), (-6.28, 0), (-3.14, 0), (0, 0), (3.14, 0), (6.28, 0), (9.42, 0)
Vertical asymptotes
none found in view
Horizontal asymptotes
none found
Domain
all real numbers in view
Range (estimated)
approximately [-1, 1]

Tip

The inverse-derivative formula evaluates f' at f⁻¹(x), never at x itself — the two arguments only coincide by accident, so always find f⁻¹(x) first (or use it symbolically) before evaluating f'.

Common Mistakes

  • Computing (f⁻¹)'(x) as 1/f'(x), evaluating f' at x instead of at f⁻¹(x).

    The correct denominator is f' evaluated at f⁻¹(x), not at x — in the worked example, that meant evaluating f'(1), not f'(3), since f⁻¹(3) = 1.

  • Confusing 1/f'(x) — the reciprocal of a function's own derivative — with (f⁻¹)'(x), the derivative of a genuinely different function.

    These agree only at the specific matched inputs the inverse relationship connects (x and f⁻¹(x)) — they are not the same expression evaluated at the same input in general.

Key Takeaways

  • (f⁻¹)'(x) = 1/f'(f⁻¹(x)) follows directly from differentiating both sides of f(f⁻¹(x)) = x with the chain rule.
  • This lets an inverse function's derivative be found at a point without ever solving for the inverse function explicitly.
  • The derivatives of arcsin and arctan are both derived the same way — implicitly differentiating their defining equations, sin(y) = x and tan(y) = x.

Summary

This closes Unit 3: the chain rule now reaches composite, implicit, and inverse functions. Unit 4 puts differentiation to work in applied contexts, starting with quantities that change together over time.

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