---
title: The Inverse Function Theorem in One Variable
module: Differentiation
moduleNumber: 5
lessonNumber: 4
order: 504
summary: >
  A nonzero derivative certifies a local inverse and fixes its slope. A strictly
  monotone differentiable function has a differentiable inverse whose derivative
  is the reciprocal of the original; the inverse function theorem removes the
  monotonicity hypothesis, and the reciprocal formula constructs nth roots and
  the logarithm's derivative, failing exactly where the derivative vanishes.
topics: [Differentiation]
sources:
  - book: Lebl
    ref: "Ch. 4 — The Derivative; §4.4 Inverse function theorem"
draft: false
---

A linear function $f(x) = ax$ with $a \neq 0$ is a bijection with inverse
$f^{-1}(y) = \frac{1}{a}y$; its derivative is $a$ and its inverse's derivative is
$\frac{1}{a}$. Since a differentiable function is infinitesimally linear, the same
reciprocal relationship should govern the inverse of any differentiable function
where the derivative is nonzero. The inverse function theorem makes this exact and
local: a single nonzero value $f'(x_0)$ guarantees an inverse on a neighborhood and
computes its derivative.

## The derivative of an inverse

A function already known to be invertible has a differentiable inverse wherever
its derivative is nonzero, and the two slopes are reciprocal.

> **Lemma (Derivative of the inverse).** Let $I, J \subseteq \mathbb{R}$ be
> intervals and $f : I \to J$ be strictly monotone and onto (hence a bijection),
> differentiable at $x_0 \in I$ with $f'(x_0) \neq 0$. Then the inverse $f^{-1}$
> is differentiable at $y_0 = f(x_0)$ and
> $$
> \bigl(f^{-1}\bigr)'(y_0) = \frac{1}{f'(x_0)} = \frac{1}{f'\bigl(f^{-1}(y_0)\bigr)}.
> $$
> If $f$ is continuously differentiable and $f'$ is never zero, then $f^{-1}$ is
> continuously differentiable.[^lebl-lemma]

> **Proof.** A strictly monotone function on an interval has a continuous
> inverse; call it $g = f^{-1} : J \to I$. For $x \neq x_0$, writing $y = f(x)$ so
> $y \neq y_0$, the inverse's difference quotient is the reciprocal of the
> original's:
> $$
> \frac{g(y) - g(y_0)}{y - y_0} = \frac{x - x_0}{f(x) - f(x_0)}.
> $$
> Define
> $$
> Q(x) =
> \begin{cases}
> \dfrac{x - x_0}{f(x) - f(x_0)}, & x \neq x_0,\\[8pt]
> \dfrac{1}{f'(x_0)}, & x = x_0,
> \end{cases}
> $$
> which is legitimate because $f'(x_0) \neq 0$. Differentiability of $f$ at $x_0$
> makes $Q$ continuous at $x_0$. Since $g$ is continuous at $y_0$, the composition
> $Q\bigl(g(y)\bigr) = \frac{g(y) - g(y_0)}{y - y_0}$ is continuous at $y_0$, so
> $$
> \lim_{y \to y_0} \frac{g(y) - g(y_0)}{y - y_0}
> = Q\bigl(g(y_0)\bigr) = \frac{1}{f'(x_0)}.
> $$
> Thus $g$ is differentiable at $y_0$ with the stated derivative. When $f'$ is
> continuous and nonzero throughout, $g'(y) = \frac{1}{f'(g(y))}$ is a composition
> of continuous functions, hence continuous. $\blacksquare$

A shorter but less self-contained derivation of the formula differentiates the
identity $f\bigl(g(y)\bigr) = y$ with the
[chain rule](/real-analysis/differentiation/the-derivative), giving
$f'\bigl(g(y)\bigr)\,g'(y) = 1$; the lemma is the honest version that first proves
$g$ is differentiable rather than assuming it.

$$
% caption: A function and its inverse are mirror images across the line y = x, so
% a tangent of slope m at a point becomes a tangent of slope one over m at the
% mirrored point.
\begin{tikzpicture}[scale=1.15,>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,->] (-0.2,0) -- (3.7,0) node[right]{x};
\draw[black,->] (0,-0.2) -- (0,3.5) node[above]{y};
\draw[black,densely dashed] (0,0) -- (3.4,3.4);
\node[black,rotate=45] at (3.0,3.25) {y = x};
% f : steep convex increasing
\draw[thick,black] (0.3,0.15) to[out=35,in=210] (1.5,1.0) to[out=45,in=225] (2.8,2.7);
\node[black] at (2.55,2.15) {f};
% inverse : mirror image (shallow)
\draw[thick,black] (0.15,0.3) to[out=55,in=245] (1.0,1.5) to[out=65,in=215] (2.7,2.8);
\node[black] at (1.15,2.35) {inverse};
% points
\fill[black] (1.5,1.0) circle (1.6pt);
\fill[black] (1.0,1.5) circle (1.6pt);
\node[below right,black] at (1.5,1.0) {A};
\node[above left,black] at (1.0,1.5) {A};
% tangent segments (reciprocal slopes)
\draw[acc] (1.0,0.3) -- (2.0,1.7);
\draw[acc] (0.3,1.05) -- (1.7,1.95);
\end{tikzpicture}
$$

The graph of $f^{-1}$ is the graph of $f$
mirrored across the diagonal $y = x$. Mirroring swaps the roles of run and rise,
so a tangent of slope $m$ becomes a tangent of slope $\frac{1}{m}$, which is the
reciprocal formula read geometrically.

## The inverse function theorem

The lemma assumes global monotonicity. The theorem removes that assumption: a
single nonzero derivative _produces_ a monotone piece, using the results of the
[mean value theorem](/real-analysis/differentiation/mean-value-theorem).

> **Theorem (Inverse function theorem).** Let $f : (a, b) \to \mathbb{R}$ be
> continuously differentiable and let $x_0 \in (a, b)$ satisfy $f'(x_0) \neq 0$.
> Then there is an open interval $I \subseteq (a, b)$ containing $x_0$ on which
> $f$ is injective, its image $J = f(I)$ is an interval, and the inverse
> $g : J \to I$ is continuously differentiable with[^lebl-thm]
> $$
> g'(y) = \frac{1}{f'\bigl(g(y)\bigr)}, \qquad y \in J.
> $$

> **Proof.** Say $f'(x_0) > 0$ (otherwise use $-f$). Continuity of $f'$ gives an
> open interval $I = (x_0 - \delta, x_0 + \delta)$ on which $f' > 0$. By the
> monotonicity test, $f$ is strictly increasing on $I$, hence injective, and its
> restriction is a bijection onto $J = f(I)$. The
> [intermediate value theorem](/real-analysis/continuity/evt-ivt) makes $J$ an
> interval. Now the lemma applies on $I$. $\blacksquare$

The theorem is genuinely local. Where the derivative changes sign, the interval
$I$ cannot be extended across the sign change even though $f'(x_0) \neq 0$.

$$
% caption: A nonzero derivative at x0 forces the same sign nearby, so f is
% strictly monotone on a small interval I and maps it bijectively onto an
% interval J where the inverse lives.
\begin{tikzpicture}[scale=1.1,>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,->] (-0.2,0) -- (5.4,0) node[right]{x};
\draw[black,->] (0,-0.2) -- (0,3.6) node[above]{y};
\draw[thick,black] (0.5,0.5) to[out=35,in=200] (2.6,2.0) to[out=20,in=210] (4.8,3.3);
% interval I on x axis
\draw[acc,very thick] (1.7,0) -- (3.4,0);
\node[acc,below] at (2.55,-0.05) {I};
% interval J on y axis
\draw[acc,very thick] (0,1.35) -- (0,2.55);
\node[acc,left] at (0,1.95) {J};
% projection lines
\draw[black,densely dashed] (1.7,0) -- (1.7,1.35) -- (0,1.35);
\draw[black,densely dashed] (3.4,0) -- (3.4,2.55) -- (0,2.55);
\fill[acc] (2.55,1.95) circle (1.6pt);
\node[above left,acc] at (2.55,1.95) {x0};
\end{tikzpicture}
$$

### Necessity of a continuous derivative

The theorem assumes $f$ is _continuously_ differentiable, not merely
differentiable, and the extra hypothesis is not decorative. A single positive
value $f'(x_0) > 0$ does not by itself force $f$ to be monotone near $x_0$; that
conclusion used continuity of $f'$ to spread the positive sign to a whole
interval.

Consider
$$
f(x) =
\begin{cases}
x + 2x^2 \sin(1/x), & x \neq 0,\\
0, & x = 0.
\end{cases}
$$
At the origin the difference quotient is $1 + 2x\sin(1/x) \to 1$, so $f'(0) = 1 > 0$.
Away from $0$,
$$
f'(x) = 1 + 4x\sin(1/x) - 2\cos(1/x),
$$
and the $-2\cos(1/x)$ term makes $f'$ oscillate between values near $-1$ and $3$ as
$x \to 0$. So $f'$ takes negative values in every neighborhood of the origin: $f$
is increasing at $0$ yet decreasing on infinitely many nearby intervals, hence not
injective on any interval around $0$ and not invertible there.[^lebl-counter] The
derivative $f'$ is discontinuous at $0$ — the very case the
continuity hypothesis of the inverse function theorem rules out.

## Worked inverses

> **Worked example ($n$th roots exist and are differentiable).** For
> $n \in \mathbb{N}$, the map $f(y) = y^n$ on $(0, \infty)$ has continuous
> derivative $f'(y) = n y^{n-1} > 0$, so it is strictly increasing and, by the
> intermediate value theorem, onto $(0, \infty)$. Its inverse $g(x) = x^{1/n}$ is
> therefore continuously differentiable, with
> $$
> g'(x) = \frac{1}{f'(g(x))} = \frac{1}{n\,(x^{1/n})^{n-1}}
> = \frac{1}{n}\,x^{(1-n)/n}.
> $$
> This is a full existence-and-smoothness proof for radicals, resting only on the
> differentiation results above, and it strengthens the bare existence of
> $\sqrt{2}$ that the completeness axiom guarantees.[^lebl-roots]

The domain restriction depends on the parity of $n$. For **even** $n$ the map
$y^n$ is not injective on $\mathbb{R}$ — it identifies $y$ with $-y$ — so the
$n$th root is defined only after restricting to $(0, \infty)$, and each positive
$x$ has one positive and one negative real $n$th root. For **odd** $n \geq 3$ the
map $y^n$ is a strictly increasing bijection of all of $\mathbb{R}$, so every real
$x$, including negatives, has a unique real $n$th root; the inverse is
differentiable everywhere except at $0$, where $f'(y) = n y^{n-1}$ vanishes and the
root inherits a vertical tangent.[^lebl-parity]

| $f$ on its domain | $f'$ | inverse $g$ | $g'(y)$ via the formula |
| --- | --- | --- | --- |
| $ax$, $a \neq 0$ | $a$ | $\tfrac{1}{a}y$ | $\tfrac{1}{a}$ |
| $y^2$ on $(0,\infty)$ | $2y$ | $\sqrt{y}$ | $\tfrac{1}{2\sqrt{y}}$ |
| $y^n$ on $(0,\infty)$ | $n y^{n-1}$ | $y^{1/n}$ | $\tfrac{1}{n} y^{(1-n)/n}$ |
| $y^3$ on $\mathbb{R}$ | $3y^2$ | $y^{1/3}$ | $\tfrac{1}{3} y^{-2/3}$ (fails at $0$) |

> **Worked example (The logarithm's derivative).** Granting the exponential from
> the [integral construction](/real-analysis/riemann-integration/log-exp-improper),
> strictly increasing on $\mathbb{R}$ with $\exp' = \exp$ never zero, the inverse
> formula computes the derivative of $\log = \exp^{-1}$ in one line:
> $$
> \log'(y) = \frac{1}{\exp'\bigl(\log y\bigr)} = \frac{1}{\exp(\log y)}
> = \frac{1}{y}, \qquad y > 0.
> $$
> The derivative of the logarithm falls out of the reciprocal rule with no
> separate computation.

**Higher smoothness transfers to the inverse.** If $f$ has $k$ continuous
derivatives near $x_0$ and $f'(x_0) \neq 0$, then so does $g = f^{-1}$.
Differentiating $g'(y) = \frac{1}{f'(g(y))}$ by the chain and quotient rules
expresses $g''$ through $f''$, $f'$, and $g'$, all continuous; each further
differentiation stays within continuous functions, so membership in $C^k$ passes
from $f$ to its inverse.[^lebl-ck]

**The theorem is only local for $y^2$.** Take $f(x) = x^2$ on all of
$\mathbb{R}$. Then $f'(x_0) \neq 0$ exactly when $x_0 \neq 0$. For $x_0 > 0$ the
largest interval on which $f$ is injective is $I = (0, \infty)$, strictly smaller
than the domain: no global inverse exists because $f$ is not
globally injective.

$$
% caption: The square is two-to-one on the line, so a horizontal level meets it
% twice; restricting to x > 0 leaves a strictly increasing branch that the
% inverse function theorem inverts, and no larger interval works.
\begin{tikzpicture}[scale=1.0,>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,->] (-2.6,0) -- (2.7,0) node[right]{x};
\draw[black,->] (0,-0.3) -- (0,3.2) node[above]{y};
\draw[thick,black] plot[domain=-1.7:0,samples=40] (\x,{\x*\x});
\draw[thick,acc] plot[domain=0:1.7,samples=40] (\x,{\x*\x});
\draw[black,densely dashed] (-2.3,1.44) -- (2.3,1.44);
\fill[black] (-1.2,1.44) circle (1.5pt);
\fill[acc] (1.2,1.44) circle (1.5pt);
\node[acc,anchor=west] at (1.55,2.7) {branch on x > 0};
\node[black,above] at (-1.35,1.5) {two preimages};
\end{tikzpicture}
$$

> **Worked example (Failure at a vanishing derivative).** The cube
> $f(x) = x^3$ is a bijection of $\mathbb{R}$ with continuous derivative, so its
> inverse $f^{-1}(y) = y^{1/3}$ exists on the entire line. But $f'(0) = 0$, and
> there the reciprocal formula would divide by zero. Correspondingly, $y^{1/3}$
> has a vertical tangent at the origin and is not differentiable there, even
> though it is continuous and a genuine inverse everywhere.

$$
% caption: The cube is a bijection with a horizontal tangent at the origin, so
% its inverse the cube root has a vertical tangent there and no derivative, the
% one point where the reciprocal formula breaks.
\begin{tikzpicture}[scale=1.0,>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,->] (-2.6,0) -- (2.7,0) node[right]{x};
\draw[black,->] (0,-2.4) -- (0,2.6) node[above]{y};
\draw[black,densely dashed] (-2.2,-2.2) -- (2.2,2.2);
% cube : flat at origin
\draw[thick,black] (-1.75,-2.1) to[out=20,in=200] (0,0) to[out=20,in=200] (1.75,2.1);
\node[black] at (1.55,1.35) {cube};
% cube root : steep at origin
\draw[thick,black] (-2.1,-1.75) to[out=70,in=250] (0,0) to[out=70,in=250] (2.1,1.75);
\node[black] at (1.15,1.9) {cube root};
\fill[acc] (0,0) circle (1.6pt);
\node[below right,acc] at (0.05,-0.1) {vertical tangent};
\end{tikzpicture}
$$

[^lebl-lemma]: **Lebl**, _Basic Analysis I_, §4.4, Lemma 4.4.1 — differentiability of the inverse and the reciprocal-derivative formula.
[^lebl-thm]: **Lebl**, _Basic Analysis I_, §4.4, Theorem 4.4.2 — the inverse function theorem in one variable via strict monotonicity from a nonzero derivative.
[^lebl-roots]: **Lebl**, _Basic Analysis I_, §4.4, Corollary 4.4.3 and Examples 4.4.4–4.4.5 — existence and differentiability of $n$th roots, and the failure of differentiability of the cube root at the origin.
[^lebl-counter]: **Lebl**, _Basic Analysis I_, §4.4, Exercise 4.4.6 — a function differentiable everywhere with $f'(0) > 0$ that is invertible on no neighborhood of the origin.
[^lebl-ck]: **Lebl**, _Basic Analysis I_, §4.4, Exercise 4.4.5 — the inverse inherits $k$ continuous derivatives from $f$.
[^lebl-parity]: **Lebl**, _Basic Analysis I_, §4.4, Exercises 4.4.3–4.4.4 — existence and differentiability of even and odd real $n$th roots.