= Solution
{c}
The <inverse function theorem> states that if $f$ is continuously differentiable near $a\in\mathbb R^n$ and $Df(a)$ is invertible, then neighborhoods of $a$ and $f(a)$ exist on which $f$ is a bijection with a continuously differentiable inverse, satisfying $D(f^{-1})(f(a))=Df(a)^{-1}$.
Here $a=(\pi/2,\pi)$, $f(a)=(0,0)$, and
$$
Df(a)=\begin{pmatrix}0&-e^{\pi/2}\\-e^{-\pi}&0\end{pmatrix},
\qquad\det Df(a)=-e^{-\pi/2}\ne0.
$$
The hypotheses hold because both components are smooth. Inverting the matrix yields
$$
\boxed{D(f^{-1})(0,0)=
\begin{pmatrix}0&-e^\pi\\-e^{-\pi/2}&0\end{pmatrix}.}
$$
Back to article page
