The Inverse Function Theorem in One Variable

A linear function $f (x) = a x$ 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.

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) = ⎩ ⎨ ⎧ \frac{x - x _{0}}{f ( x ) - f ( x _{0} )}, \frac{1}{f ^{'} ( x _{0} )}, x \neq = x_{0}, x = x_{0},

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 $Q (g (y)) = \frac{g ( y ) - g ( y _{0} )}{y - y _{0}}$ is continuous at $y_{0}$ , so

y \to y_{0} lim \frac{g ( y ) - g ( y _{0} )}{y - y _{0}} = Q (g (y_{0})) = \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. $■$

A shorter but less self-contained derivation of the formula differentiates the identity $f (g (y)) = y$ with the chain rule, giving $f^{'} (g (y)) g^{'} (y) = 1$ ; the lemma is the honest version that first proves $g$ is differentiable rather than assuming it.

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.

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.

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

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.

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) = {x + 2 x^{2} sin (1/ x), 0, x \neq = 0, x = 0.

At the origin the difference quotient is $1 + 2 x sin (1/ x) \to 1$ , so $f^{'} (0) = 1 > 0$ . Away from $0$ ,

f^{'} (x) = 1 + 4 x 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.³ The derivative $f^{'}$ is discontinuous at $0$ — the very case the continuity hypothesis of the inverse function theorem rules out.

Worked inverses

The domain restriction depends on the parity of $n$ . For even $n$ the map $y^{n}$ is not injective on $R$ — 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 $R$ , 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.⁵

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

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.⁶

The theorem is only local for $y^{2}$ . Take $f (x) = x^{2}$ on all of $R$ . 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.

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.

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.

Lebl, Basic Analysis I, §4.4, Lemma 4.4.1 — differentiability of the inverse and the reciprocal-derivative formula. ↩
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, 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, 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, Basic Analysis I, §4.4, Exercises 4.4.3–4.4.4 — existence and differentiability of even and odd real $n$ th roots. ↩
Lebl, Basic Analysis I, §4.4, Exercise 4.4.5 — the inverse inherits $k$ continuous derivatives from $f$ . ↩