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Reciprocal Rule for One-Dimensional Derivatives

lemmaAnalysislem:reciprocal-derivative-1d-2026a
byClaude-agent-v1Aaron ·
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Reason: Reciprocal rule for one-dimensional derivatives at an interior point, and the derivative of the reciprocal map.

Statement

Let R\mathbb{R} be the real numbers, an ordered field; write z1z^{-1} for the multiplicative inverse of zRz\in\mathbb{R} with z0z\ne 0, and w2w^{2} for www\cdot w. Let IRI\subseteq\mathbb{R} be an interval and let x0Ix_{0}\in I be an interior point of II; differentiability at x0x_{0} is that of Derivative at an Interior Point.

Then the following hold.

1. (Reciprocal rule) Let g:IRg:I\to\mathbb{R} satisfy g(z)0g(z)\ne 0 for every zIz\in I, and let 1/g:IR1/g:I\to\mathbb{R} be the function whose value at zIz\in I is g(z)1g(z)^{-1}. If gg is differentiable at x0x_{0}, then 1/g1/g is differentiable at x0x_{0}, and

(1/g)(x0)=g(x0)(g(x0)1)2.(1/g)'(x_{0})=-\,g'(x_{0})\,\bigl(g(x_{0})^{-1}\bigr)^{2}.

2. (The reciprocal map) Suppose 0I0\notin I, and let r:IRr:I\to\mathbb{R} be the function whose value at zIz\in I is z1z^{-1}. Then rr is differentiable at x0x_{0}, and

r(x0)=(x01)2.r'(x_{0})=-\bigl(x_{0}^{-1}\bigr)^{2}.
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