Question

Theorem 7.2.1 Derivatives of Inverse Functions Let $f$ be differentiable and one-to-one on an open interval $I$, where $f'(x) \neq 0$ for all $x$ in $I$, let $J$ be the range of $f$ on $I$, let $g$ be the inverse function of $f$, and let $f(a) = b$ for some $a$ in $I$. Then $g$ is a differentiable function on $J$, and in particular, $(f^{-1})'(b) = g'(b) = \frac{1}{f'(a)}$ $(f^{-1})'(x) = g'(x) = \frac{1}{f'(g(x))}$

          Theorem 7.2.1
Derivatives of Inverse Functions
Let $f$ be differentiable and one-to-one on an open interval $I$, where $f'(x) \neq 0$ for all $x$ in $I$, let $J$ be the range of $f$ on
$I$, let $g$ be the inverse function of $f$, and let $f(a) = b$ for some $a$ in $I$. Then $g$ is a differentiable function on $J$, and in
particular,
$(f^{-1})'(b) = g'(b) = \frac{1}{f'(a)}$
$(f^{-1})'(x) = g'(x) = \frac{1}{f'(g(x))}$

Theorem 7.2.1
Derivatives of Inverse Functions
Let f be differentiable and one-to-one on an open interval I, where f'(x) ≠ 0 for all x in I, let J be the range of f on
I, let g be the inverse function of f, and let f(a) = b for some a in I. Then g is a differentiable function on J, and in
particular,
(f^-1)'(b) = g'(b) = (1)/(f'(a))
(f^-1)'(x) = g'(x) = (1)/(f'(g(x)))

Added by Donald R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ben Blakesley

03/08/2022

Step 1

Step 1: Since f is differentiable and one-to-one on the open interval I, and f'(x) ≠ 0 for all x in I, we know that f has an inverse function g. Show more…

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Theorem 7.2.1 Derivatives of Inverse Functions Let f be differentiable and one-to-one on an open interval I, where f'(x) ≠ 0 for all x in I, let J be the range of f on I, let g be the inverse function of f, and let f(a) = b for some a in I. Then g is a differentiable function on J, and in particular, (f^(-1))'(b) = g'(b) = (1) / (f'(a)) (f^(-1))'(x) = g'(x) = (1) / (f'(g(x)))