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Inverse Function Theorem

Theorem

Let XX and YY be open intervals in R\mathbb{R}, and let f ⁣:X→Yf\colon X\to Y be an invertible function, with inverse f−1 ⁣:Y→X.f^{-1}\colon Y\to X\text{.} Suppose that x∈Xx\in X and y∈Yy\in Y are such that f(x)=y,f(x)=y\text{,} or equivalently, x=f−1(y).x=f^{-1}(y)\text{.} If ff is differentiable at x,x\text{,} f′(x)≠0,f'(x)\neq 0\text{,} and f−1f^{-1} is continuous at y,y\text{,} then f−1f^{-1} is differentiable at yy and

(f−1)′(y)=1f′(x)=1f′(f−1(y)).\left(f^{-1}\right)'(y)=\frac{1}{f'(x)}=\frac{1}{f'(f^{-1}(y))}.

Interpretation

The inverse function theorem says when the inverse of a differentiable function is again differentiable, and how to find its derivative without a formula for f−1f^{-1}: the slope of f−1f^{-1} at yy is the reciprocal of the slope of ff at the matching point x=f−1(y)x=f^{-1}(y).

The graphs of x squared and its inverse, the square root, mirrored in the line y = x; the tangent at x0 has slope f'(x0) and the mirrored tangent at y0 has slope 1/f'(x0).

The graph of f−1f^{-1} is the graph of ff reflected in the line y=xy=x, and the reflection swaps rise and run. Here f(x)=x2f(x)=x^2 and f−1(y)=yf^{-1}(y)=\sqrt{y}: the tangent to ff at (x0,y0)(x_0,y_0) has slope f′(x0)=2x0f'(x_0)=2x_0, and its mirror image, the tangent to f−1f^{-1} at (y0,x0)(y_0,x_0), has slope 1/(2x0)1/(2x_0), which is indeed the derivative of y\sqrt{y} at y0=x02y_0=x_0^2.

Each hypothesis has a role. If f′(x)=0f'(x)=0, the tangent of ff is horizontal and its mirror image is vertical, so f−1f^{-1} has no derivative at yy: f(x)=x3f(x)=x^3 has f′(0)=0f'(0)=0, and the cube root has a vertical tangent at 00. Continuity of f−1f^{-1} at yy makes f−1(y+k)→xf^{-1}(y+k)\to x as k→0k\to0, so that difference quotients of f−1f^{-1} are reciprocals of difference quotients of ff near xx. When ff is continuous on the whole interval XX, it is strictly monotone, and then f−1f^{-1} is automatically continuous.

The formula is easy to remember through the chain rule: differentiating f(f−1(y))=yf(f^{-1}(y))=y gives f′(f−1(y)) (f−1)′(y)=1f'(f^{-1}(y))\,(f^{-1})'(y)=1. This calculation assumes that (f−1)′(y)(f^{-1})'(y) exists; the theorem guarantees it. A common slip is to evaluate f′f' at the wrong point: (f−1)′(y)(f^{-1})'(y) is 1/f′(x)1/f'(x) with x=f−1(y)x=f^{-1}(y), not 1/f′(y)1/f'(y). For f(x)=exf(x)=\mathrm{e}^x this gives (ln⁡)′(y)=1/eln⁡y=1/y(\ln)'(y)=1/\mathrm{e}^{\ln y}=1/y, not 1/ey1/\mathrm{e}^{y}. Math IV extends the theorem to functions of several variables in the multivariable inverse function theorem.

Key Results

  • If f′f' is nonzero throughout an open interval, then ff is strictly monotone there and hence invertible onto its range. The theorem itself assumes that ff is invertible and only concerns the differentiability of f−1.f^{-1}\text{.}
  • Differentiability of the inverse depends on continuity assumptions.
  • The derivative of the inverse is determined by the reciprocal of the original derivative.

Examples

  • If f′(x0)=2,f'(x_0)=2\text{,} then the derivative of the inverse is 1/21/2 at y=f(x0).y=f(x_0)\text{.}
  • The derivative ex\mathrm{e}^x at x=1/2x=1/2 is the reciprocal of the derivative of ln⁡y\ln y at y=e1/2.y=\mathrm{e} ^{1/2}\text{.}

Prerequisites

Sources

Aliases: inverse function rule, derivative of the inverse function Symbols: element of, real numbers, square root

Last updated: 20 October 2026 03:02