Inverse function rule ($(f^{-1}{)}^{\prime }(b)=1/f^{\prime }(a),b=f(a)$)

Suppose $f$ is continuously differentiable near $a$ with $f^{\prime }(a)\ne 0$. It has a local inverse function $g$ near $b=f(a)$, andThe 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. ^[ref].