Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28
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Derivative of an inverse: if ff is continuous and injective on a nondegenerate interval II and differentiable at cIc \in I with f(c)0f'(c) \ne 0, then the inverse gg is differentiable at f(c)f(c) with g(f(c))=1/f(c)g'(f(c)) = 1/f'(c); and if f(c)=0f'(c) = 0 then gg is not differentiable at f(c)f(c)

Statement

Let IRI \subseteq \mathbb{R} be order-convex with at least two elements (Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length), let f:IRf : I \to \mathbb{R} be continuous on II (Continuity of f:ARf : A \to \mathbb{R} at a point of AA and on AA: the ε\varepsilon-δ\delta condition, its agreement with limxcf(x)=f(c)\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]Ig : f[I] \to I be the inverse of f:If[I]f : I \to f[I] supplied by Continuous inverse theorem: a continuous injective ff on an interval II is a bijection onto the order-convex set f[I]f[I], and the inverse g:f[I]Ig : f[I] \to I is continuous and strictly monotone in the same sense as ff. Let cIc \in I and put b:=f(c)b := f(c).

Then cc is a limit point of II and bb is a limit point of f[I]f[I], so that f(c)f'(c) and g(b)g'(b) are meaningful symbols (The derivative f(c)=limxcf(x)f(c)xcf'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} of f:ARf : A \to \mathbb{R} at a point cAc \in A that is a limit point of AA, and differentiability on a set), and, assuming ff is differentiable at cc:

  1. if f(c)0f'(c) \ne 0, then gg is differentiable at bb and g(b)  =  1f(c);g'(b) \;=\; \frac{1}{f'(c)} ;
  2. if f(c)=0f'(c) = 0, then gg is not differentiable at bb.

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]f[I] that is not of the form f(c)f(c) with ff differentiable at cc, and nothing is asserted about gg being differentiable on a set.

No compactness and no boundedness is assumed. II 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 II a limit point of II (The derivative f(c)=limxcf(x)f(c)xcf'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} of f:ARf : A \to \mathbb{R} at a point cAc \in A that is a limit point of AA, and differentiability on a set).

Facts & Assumptions

[L1]

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

[L2]

Carathéodory's characterisation (Carathéodory's characterisation: ff is differentiable at cc if and only if there is φ:AR\varphi : A \to \mathbb{R}, continuous at cc, with f(x)f(c)=φ(x)(xc)f(x) - f(c) = \varphi(x)(x - c) for every xAx \in A, and then φ\varphi is unique and φ(c)=f(c)\varphi(c) = f'(c)), used in both directions: for DRD \subseteq \mathbb{R}, a point pDp \in D that is a limit point of DD and h:DRh : D \to \mathbb{R}, the function hh is differentiable at pp if and only if there is η:DR\eta : D \to \mathbb{R}, continuous at pp, with h(y)h(p)=η(y)(yp)h(y) - h(p) = \eta(y)(y-p) for every yDy \in D, and then η(p)=h(p)\eta(p) = h'(p).

[L4]

Injectivity (Injection, surjection, bijection): f(x)=f(x)f(x) = f(x') implies x=xx = x', so xcx \ne c gives f(x)f(c)f(x) \ne f(c); and the image f[I]={f(x):xI}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:DRu, v : D \to \mathbb{R} are continuous at pDp \in D with v(p)0v(p) \ne 0, then (u/v)(u/v) restricted to {yD:v(y)0}\{ y \in D : v(y) \ne 0 \} is continuous at pp (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 gg is differentiable at cc and ff is differentiable at g(c)g(c), then fgf \circ g is differentiable at cc with (fg)(c)=f(g(c))g(c)(f \circ g)'(c) = f'(g(c))\,g'(c)): with gg differentiable at the limit point b=f(c)b = f(c) of f[I]f[I] and ff differentiable at the limit point cc of II, the composite gfg \circ f is differentiable at cc with (gf)(c)=g(b)f(c)(g \circ f)'(c) = g'(b)\,f'(c).

Proof

technique · direct
1.1

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

L1L3L4
1.2

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

L1choose
1.3

Assume ff is differentiable at cc. By [L2], applied to ff on II at the limit point cc, fix φ:IR\varphi : I \to \mathbb{R}, continuous at cc, with f(x)f(c)=φ(x)(xc)f(x) - f(c) = \varphi(x)(x - c) for every xIx \in I and φ(c)=f(c)\varphi(c) = f'(c).

L2choose
2.1

φ(x)0\varphi(x) \ne 0 for every xIx \in I with xcx \ne c: injectivity gives f(x)f(c)f(x) \ne f(c), so φ(x)(xc)0\varphi(x)(x-c) \ne 0 and hence φ(x)0\varphi(x) \ne 0. If moreover f(c)0f'(c) \ne 0 then φ(c)=f(c)0\varphi(c) = f'(c) \ne 0 as well, so φ\varphi vanishes at no point of II.

step 1.3L4
2.2

The increment of gg, rewritten. Let uf[I]u \in f[I] and put x:=g(u)Ix := g(u) \in I, so f(x)=uf(x) = u by [L1]. Then ub=f(x)f(c)=φ(x)(xc)=φ(g(u))(g(u)g(b))u - b = f(x) - f(c) = \varphi(x)(x - c) = \varphi(g(u))\,\bigl(g(u) - g(b)\bigr), using g(b)=cg(b) = c from step 1.2.

step 1.2step 1.3L1
2.3

Claim 2. Assume f(c)=0f'(c) = 0, and suppose gg were differentiable at bb. Since f[I]f[I]f[I] \subseteq f[I], since ff is differentiable at the limit point cc of II and since b=f(c)b = f(c) is a limit point of f[I]f[I] by step 1.1, the chain rule [L6] gives that gf:IRg \circ f : I \to \mathbb{R} is differentiable at cc with (gf)(c)=g(b)f(c)=g(b)0=0(g \circ f)'(c) = g'(b)\,f'(c) = g'(b) \cdot 0 = 0. But gfg \circ f is the identity on II by step 1.2, and by [L7] the identity on II is differentiable at the limit point cc with derivative 11; the derivative at cc being a single real, this forces 0=10 = 1, which [L7] excludes. So gg is not differentiable at bb.

step 1.1step 1.2L6L7
3.1

The reciprocal factor. Assume f(c)0f'(c) \ne 0. The map gg is continuous at bb by step 1.2 and sends f[I]f[I] into II, and φ\varphi is continuous at c=g(b)c = g(b) by step 1.3, so φg:f[I]R\varphi \circ g : f[I] \to \mathbb{R} is continuous at bb by [L5]; by step 2.1 it vanishes at no point of f[I]f[I], since gg takes values in II, and (φg)(b)=φ(c)=f(c)0(\varphi \circ g)(b) = \varphi(c) = f'(c) \ne 0. Hence, by [L5] applied with the constant numerator 11 and denominator φg\varphi \circ g on the domain f[I]f[I], where the set on which the denominator does not vanish is the whole of f[I]f[I], the function Φ:=1/(φg):f[I]R\Phi := 1/(\varphi \circ g) : f[I] \to \mathbb{R} is continuous at bb and Φ(b)=1/f(c)\Phi(b) = 1/f'(c).

step 1.2step 1.3step 2.1L5
4.1

The factorisation for gg. Assume f(c)0f'(c) \ne 0 and let uf[I]u \in f[I]. Dividing the identity of step 2.2 by the nonzero number (φg)(u)(\varphi \circ g)(u) gives g(u)g(b)=Φ(u)(ub)g(u) - g(b) = \Phi(u)\,(u - b), and this holds for every uf[I]u \in f[I].

step 2.2step 3.1
5.1

Claim 1. Assume f(c)0f'(c) \ne 0. By step 1.1 the point bb is a limit point of f[I]f[I]; by step 4.1 the function Φ:f[I]R\Phi : f[I] \to \mathbb{R} factors the increment of gg at bb; and by step 3.1 it is continuous at bb. So [L2], applied to gg on f[I]f[I] at bb, gives that gg is differentiable at bb with g(b)=Φ(b)=1/f(c)g'(b) = \Phi(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=0f' = 0 no inverse can be differentiable, because the chain rule would then make the derivative of the identity equal to 00. 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 ff on an interval II is a bijection onto the order-convex set f[I]f[I], and the inverse g:f[I]Ig : f[I] \to I is continuous and strictly monotone in the same sense as ff, and what is not. Only that f[I]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))g'(b) = 1/f'(g(b)), which is the same statement since g(b)=cg(b) = c. Written that way it is a formula for gg' at every point of f[I]f[I] at which the hypothesis holds, and that is how the companion page uses it to differentiate xx1/nx \mapsto x^{1/n}.

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