= Inverse function rule
{title2=$(f^{-1})'(b)=1/f'(a),\quad b=f(a)$}
{wiki}
Suppose $f$ is <continuously differentiable> near $a$ with $f'(a)\ne0$. It has a local <inverse function> $g$ near $b=f(a)$, and
$$
g'(b)=\frac1{f'(a)}=\frac1{f'(g(b))}.
$$
The <chain rule> applied to $g(f(x))=x$ gives the formula. The nonzero <derivative> condition also ensures local invertibility; it does not assert a global <inverse function> on an arbitrary larger <domain of a function>. For $f(x)=x^2$ at $a=2$, the corresponding output is $b=4$, and the <square root>'s <derivative> there is $1/4$. At $a=0$, this division is unavailable: the nonnegative restriction still has a <square root> inverse, but its finite <derivative> at zero does not exist. https://openstax.org/books/calculus-volume-1/pages/3-7-derivatives-of-inverse-functions[OpenStax, the inverse function rule and corresponding evaluation points]{ref}.
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