Derivatives of Inverse Functions

When a function f has an inverse f⁻¹, there is a beautiful relationship between their derivatives. Rather than finding an explicit formula for the inverse and differentiating it directly — which is often algebraically painful or even impossible — we can use the Inverse Function Theorem to compute the derivative of f⁻¹ directly from the derivative of f. This powerful shortcut is one of the elegant results of calculus. The key insight is geometric: the graph of f⁻¹ is the reflection of the graph of f across the line y = x. This reflection swaps the roles of x and y, which means slopes (rise over run) become their reciprocals (run over rise). Algebraically, if f(a) = b, then f⁻¹(b) = a, and the derivative of f⁻¹ at b equals 1 divided by the derivative of f at a. This reciprocal relationship is the heart of the theorem. Derivatives of inverse functions appear constantly in calculus: deriving the formulas for the derivatives of arcsin, arccos, arctan, logarithms, and other inverse functions all rely on this theorem. Mastering this concept gives you the tools to differentiate a vast new family of functions without memorising each formula from scratch.

Key takeaways
  • Start with the definition of inverse functions
  • Differentiate both sides using the Chain Rule
  • Solve for the derivative of the inverse
  • Alternative notation using a and b
  • Classic application: derivative of arctan(x)
  • When does the theorem fail?

Start with the definition of inverse functions

If f and f⁻¹ are inverse functions, then by definition f(f⁻¹(x)) = x for every x in the domain of f⁻¹. This composition identity is the starting point for our derivation.

Differentiate both sides using the Chain Rule

Differentiating both sides of f(f⁻¹(x)) = x with respect to x gives: f'(f⁻¹(x)) · (f⁻¹)'(x) = 1. The left side uses the Chain Rule — the derivative of the outer function f evaluated at the inner function f⁻¹(x), multiplied by the derivative of the inner function (f⁻¹)'(x). The right side, the derivative of x, is simply 1.

Solve for the derivative of the inverse

Dividing both sides by f'(f⁻¹(x)), provided this value is not zero, yields the Inverse Function Theorem formula: (f⁻¹)'(x) = 1 / f'(f⁻¹(x)). This is valid wherever f'(f⁻¹(x)) ≠ 0, which geometrically means the tangent to f at the corresponding point is not horizontal.

Alternative notation using a and b

If f(a) = b (equivalently f⁻¹(b) = a), then the theorem states (f⁻¹)'(b) = 1 / f'(a). This form is especially useful for evaluating the derivative of an inverse at a specific point without finding a general formula for f⁻¹.

Classic application: derivative of arctan(x)

Let f(x) = tan(x) so f⁻¹(x) = arctan(x). We know f'(x) = sec²(x). Using the theorem: d/dx[arctan(x)] = 1 / f'(arctan(x)) = 1 / sec²(arctan(x)). Now set θ = arctan(x), so tan(θ) = x. Using the identity sec²(θ) = 1 + tan²(θ) = 1 + x², we get d/dx[arctan(x)] = 1/(1 + x²). This is the standard result, derived entirely from the theorem.

When does the theorem fail?

The Inverse Function Theorem requires f'(a) ≠ 0 at the corresponding point. If f'(a) = 0, the tangent to f is horizontal, and the tangent to f⁻¹ would need to be vertical (undefined slope). The theorem also requires f to be differentiable and locally one-to-one (strictly monotone) near the point of interest.

The Key Equation

(f−1)′(x)=1f′ ⁣(f−1(x))\left(f^{-1}\right)'(x) = \frac{1}{f'\!\left(f^{-1}(x)\right)}
  • f^{-1} — Inverse function (dimensionless): The inverse of f; it satisfies f(f⁻¹(x)) = x and f⁻¹(f(x)) = x.
  • (f^{-1})'(x) — Derivative of the inverse (dimensionless): The rate of change of the inverse function at the point x; this is what we are computing.
  • f'(f^{-1}(x)) — Derivative of f at the reflected point (dimensionless): The derivative of the original function evaluated at f⁻¹(x); this must be nonzero for the formula to apply.

Worked Example

Problem. Let f(x) = x³ + 2x. It can be shown that f is one-to-one. (a) Find (f⁻¹)'(3). (b) Find d/dx[arcsin(x)] using f(x) = sin(x).

(a)  (f−1)′(3)=15;(b)  ddxarcsin⁡(x)=11−x2\text{(a)}\;\left(f^{-1}\right)'(3)=\frac{1}{5};\qquad \text{(b)}\;\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}

Answer. Part (a): 1/5. Part (b): d/dx[arcsin(x)] = 1/√(1-x²)

Figure
Diagram
Inverse Functions and Their Tangent Linesxyy = xf(x)f⁻¹(x)(a, b)(b, a)slope = f′(a)slope = 1/f′(a)Tangent slopes at reflected points are reciprocals of each other
Origins · 1 minGottfried Wilhelm Leibniz and how this was discovered
GW
Gottfried Wilhelm Leibniz
1646–1716 · German

Leibniz co-invented calculus independently of Newton and developed much of the notation we use today, including dy/dx for derivatives and the integral sign ∫. His systematic approach to differentiating composed and inverse functions laid the conceptual groundwork for results like the Inverse Function Theorem. The Chain Rule, which is essential for deriving the inverse function derivative formula, was made explicit in his framework. Leibniz was also a philosopher and diplomat who designed an early mechanical calculator called the Stepped Reckoner. He could read at age 6 and taught himself Latin at 8 to access scholarly books in his father's library.

  • The Chain Rule — The derivation of the Inverse Function Theorem uses the Chain Rule as its central step when differentiating f(f⁻¹(x)) = x.
  • Implicit Differentiation — An alternative method for differentiating inverse functions; writing y = f⁻¹(x) as f(y) = x and differentiating implicitly gives the same result.
  • Derivatives of Inverse Trigonometric Functions — All six inverse trig derivatives (arcsin, arccos, arctan, etc.) are derived directly from the Inverse Function Theorem.
  • Derivative of the Natural Logarithm — Since ln(x) is the inverse of eˣ, the Inverse Function Theorem immediately gives d/dx[ln(x)] = 1/x.