TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$

Statement

Let $I \subseteq R$ be order-convex with at least two elements (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), let $f : I \to R$ be continuous on $I$ (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point) and injective (Injection, surjection, bijection), and let $g : f [I] \to I$ be the inverse of $f : I \to f [I]$ supplied by Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ . Let $c \in I$ and put $b : = f (c)$ .

Then $c$ is a limit point of $I$ and $b$ is a limit point of $f [I]$ , so that $f^{'} (c)$ and $g^{'} (b)$ are meaningful symbols (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set), and, assuming $f$ is differentiable at $c$ :

if $f^{'} (c) \neq 0$ , then $g$ is differentiable at $b$ and $g^{'} (b) = \frac{1}{f^{'} (c)};$
if $f^{'} (c) = 0$ , then $g$ is not differentiable at $b$ .

The two claims together say that the inverse inherits differentiability exactly where the derivative does not vanish. Nothing is asserted at a point of $f [I]$ that is not of the form $f (c)$ with $f$ differentiable at $c$ , and nothing is asserted about $g$ being differentiable on a set.

No compactness and no boundedness is assumed. $I$ may be open, half-open or unbounded; all that is used of it is order-convexity and the presence of two distinct points, the latter being exactly what makes every point of $I$ a limit point of $I$ (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

Facts & Assumptions

Given: An order-convex $I \subseteq R$ with at least two elements, a continuous injective $f : I \to R$ , a point $c \in I$ , and $b : = f (c)$ ; from step 1.3 onwards also the hypothesis that $f$ is differentiable at $c$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, Injection, surjection, bijection, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[L1]

Continuous inverse theorem (Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ , claims 2, 3 and 5): $f [I]$ is order-convex; $f : I \to f [I]$ is a bijection, so there is exactly one $g : f [I] \to I$ with $g (f (x)) = x$ for every $x \in I$ and $f (g (u)) = u$ for every $u \in f [I]$ ; and $g$ is continuous on $f [I]$ .

[L2]

Carathéodory's characterisation (Carathéodory's characterisation: $f$ is differentiable at $c$ if and only if there is $φ : A \to R$ , continuous at $c$ , with $f (x) - f (c) = φ (x) (x - c)$ for every $x \in A$ , and then $φ$ is unique and $φ (c) = f^{'} (c)$ ), used in both directions: for $D \subseteq R$ , a point $p \in D$ that is a limit point of $D$ and $h : D \to R$ , the function $h$ is differentiable at $p$ if and only if there is $η : D \to R$ , continuous at $p$ , with $h (y) - h (p) = η (y) (y - p)$ for every $y \in D$ , and then $η (p) = h^{'} (p)$ .

[L3]

Every point of an order-convex subset of $R$ with at least two elements is a limit point of that set (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ).

[L4]

Injectivity (Injection, surjection, bijection): $f (x) = f (x^{'})$ implies $x = x^{'}$ , so $x \neq c$ gives $f (x) \neq f (c)$ ; and the image $f [I] = {f (x) : x \in I}$ .

[L5]

Algebra and composition of continuous functions: a composite of functions continuous at the relevant points is continuous (A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs); every constant function is continuous (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 5); and if $u, v : D \to R$ are continuous at $p \in D$ with $v (p) \neq 0$ , then $(u / v)$ restricted to ${y \in D : v (y) \neq 0}$ is continuous at $p$ (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 4).

[L6]

Chain rule (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ): with $g$ differentiable at the limit point $b = f (c)$ of $f [I]$ and $f$ differentiable at the limit point $c$ of $I$ , the composite $g \circ f$ is differentiable at $c$ with $(g \circ f)^{'} (c) = g^{'} (b) f^{'} (c)$ .

[L7]

The identity on a set $D$ is differentiable at every limit point $p$ of $D$ with derivative $1$ : its difference quotient is $(y - p) / (y - p) = 1$ for every $y \in D ∖ {p}$ , a constant function, whose limit at $p$ is $1$ (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ). The derivative at a point is a single real (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set), and $0 \neq 1$ in $R$ (The multiplicative identity is positive).

Proof

technique · direct

1.1

$I$ has at least two elements, so by [L4] its image $f [I]$ has at least two elements; and $f [I]$ is order-convex by [L1]. So [L3] applies to both sets: every point of $I$ is a limit point of $I$ , and every point of $f [I]$ is a limit point of $f [I]$ . In particular $c$ is a limit point of $I$ and $b = f (c) \in f [I]$ is a limit point of $f [I]$ .

L1L3L4

1.2

Fix the inverse $g : f [I] \to I$ of $f : I \to f [I]$ , continuous on $f [I]$ ; it satisfies $g (f (x)) = x$ for every $x \in I$ , so in particular $g (b) = c$ .

L1choose

1.3

Assume $f$ is differentiable at $c$ . By [L2], applied to $f$ on $I$ at the limit point $c$ , fix $φ : I \to R$ , continuous at $c$ , with $f (x) - f (c) = φ (x) (x - c)$ for every $x \in I$ and $φ (c) = f^{'} (c)$ .

L2choose

2.1

$φ (x) \neq 0$ for every $x \in I$ with $x \neq c$ : injectivity gives $f (x) \neq f (c)$ , so $φ (x) (x - c) \neq 0$ and hence $φ (x) \neq 0$ . If moreover $f^{'} (c) \neq 0$ then $φ (c) = f^{'} (c) \neq 0$ as well, so $φ$ vanishes at no point of $I$ .

step 1.3L4

2.2

The increment of $g$ , rewritten. Let $u \in f [I]$ and put $x : = g (u) \in I$ , so $f (x) = u$ by [L1]. Then $u - b = f (x) - f (c) = φ (x) (x - c) = φ (g (u)) (g (u) - g (b))$ , using $g (b) = c$ from step 1.2.

step 1.2step 1.3L1

2.3

Claim 2. Assume $f^{'} (c) = 0$ , and suppose $g$ were differentiable at $b$ . Since $f [I] \subseteq f [I]$ , since $f$ is differentiable at the limit point $c$ of $I$ and since $b = f (c)$ is a limit point of $f [I]$ by step 1.1, the chain rule [L6] gives that $g \circ f : I \to R$ is differentiable at $c$ with $(g \circ f)^{'} (c) = g^{'} (b) f^{'} (c) = g^{'} (b) \cdot 0 = 0$ . But $g \circ f$ is the identity on $I$ by step 1.2, and by [L7] the identity on $I$ is differentiable at the limit point $c$ with derivative $1$ ; the derivative at $c$ being a single real, this forces $0 = 1$ , which [L7] excludes. So $g$ is not differentiable at $b$ .

step 1.1step 1.2L6L7

3.1

The reciprocal factor. Assume $f^{'} (c) \neq 0$ . The map $g$ is continuous at $b$ by step 1.2 and sends $f [I]$ into $I$ , and $φ$ is continuous at $c = g (b)$ by step 1.3, so $φ \circ g : f [I] \to R$ is continuous at $b$ by [L5]; by step 2.1 it vanishes at no point of $f [I]$ , since $g$ takes values in $I$ , and $(φ \circ g) (b) = φ (c) = f^{'} (c) \neq 0$ . Hence, by [L5] applied with the constant numerator $1$ and denominator $φ \circ g$ on the domain $f [I]$ , where the set on which the denominator does not vanish is the whole of $f [I]$ , the function $Φ : = 1 / (φ \circ g) : f [I] \to R$ is continuous at $b$ and $Φ (b) = 1 / f^{'} (c)$ .

step 1.2step 1.3step 2.1L5

4.1

The factorisation for $g$ . Assume $f^{'} (c) \neq 0$ and let $u \in f [I]$ . Dividing the identity of step 2.2 by the nonzero number $(φ \circ g) (u)$ gives $g (u) - g (b) = Φ (u) (u - b)$ , and this holds for every $u \in f [I]$ .

step 2.2step 3.1

5.1

Claim 1. Assume $f^{'} (c) \neq 0$ . By step 1.1 the point $b$ is a limit point of $f [I]$ ; by step 4.1 the function $Φ : f [I] \to R$ factors the increment of $g$ at $b$ ; and by step 3.1 it is continuous at $b$ . So [L2], applied to $g$ on $f [I]$ at $b$ , gives that $g$ is differentiable at $b$ with $g^{'} (b) = Φ (b) = 1 / f^{'} (c)$ .

step 1.1step 3.1step 4.1L2

6.1

Claim 1 is step 5.1 and claim 2 is step 2.3, and the two limit-point assertions are step 1.1.

step 2.3step 5.1∎

Remarks

Why claim 2 is not a defect of the method. It is a theorem: at a point where $f^{'} = 0$ no inverse can be differentiable, because the chain rule would then make the derivative of the identity equal to $0$ . The geometry is the familiar one, a horizontal tangent reflecting into a vertical one, and the argument above is that picture with no picture in it.
What is used of Continuous inverse theorem: a continuous injective $f$ on an interval $I$ is a bijection onto the order-convex set $f [I]$ , and the inverse $g : f [I] \to I$ is continuous and strictly monotone in the same sense as $f$ , and what is not. Only that $f [I]$ is order-convex, that the two-sided inverse exists and is unique, and that it is continuous. The strict monotonicity that theorem also proves is not needed here, though it is what makes the situation intelligible.
The formula is often written $g^{'} (b) = 1 / f^{'} (g (b))$ , which is the same statement since $g (b) = c$ . Written that way it is a formula for $g^{'}$ at every point of $f [I]$ at which the hypothesis holds, and that is how the companion page uses it to differentiate $x \mapsto x^{1 / n}$ .

Depends on

Used by

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Sources

Inverse function rule (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 5 (standard reference, not scraped)
J. Lebl, Basic Analysis I, §4.4 (standard reference, not scraped)
J. Lebl, Basic Analysis I, The Derivative (standard reference, not scraped)
J. Hunter, An Introduction to Real Analysis (standard reference, not scraped)