Answer
We find the relation
$g'[f(x)]\cdot f'(x)=1$
Work Step by Step
$(g\circ f)(x)=x$
$g[f(x)]=x$
... differentiating both sides
$\displaystyle \frac{d}{dx}\{g[f(x)]\}=\frac{d}{dx}[x]$
... using the chain rule
$g'[f(x)]\cdot f'(x)=1$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 7: Transcendental Functions - Section 7.1 - Inverse Functions and Their Derivatives - Exercises 7.1 - Page 374: 59
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