Let $f:U\subseteq \mathbb{R}^2 \rightarrow \mathbb{R}^3$ and $g: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be defined by: $$f(x,y)=\begin{pmatrix} xy \\ ye^x \\ \log x \end{pmatrix} \text{ and }\ g(x,y)=\begin{pmatrix} \sin x \\ \cos2y \\ 1 \end{pmatrix}$$, where $U$ is a suitable subset so that the function and their derivatives can be defined.
The value of $D(f \cdot g)(\pi, \pi)$ has the form $(x,y) \in \mathbb{R}^2$. Compute $x+y$.
My Attempt:
I found $Dg(x,y) = \begin{pmatrix} \cos x & 0 \\ 0 & -2\cos2y \\ 0 & 0 \end{pmatrix} \text{ and }\ Df(x,y)=\begin{pmatrix} y & x \\ ye^x & -e^x \\ \frac{1}{x} & 0 \end{pmatrix}$.
Then, $D(f\cdot g)(\pi, \pi)=(f(\pi,\pi)^TDg(\pi,\pi) \ , \ g(\pi, \pi)^TDf(\pi,\pi))$
So, $=\begin{pmatrix}\pi^2 & \pi e^\pi & \log\pi\end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix} \ , \ \begin{pmatrix} 0&1&1\end{pmatrix} \begin{pmatrix} \pi & \pi \\ \pi e^\pi & e^\pi \\ \frac{1}{\pi} & 0 \end{pmatrix}$
$=(-\pi^2, \pi e^\pi + \frac{1}{\pi}+e^\pi)$
Hence, $x+y \approx 86.288$
I was unsure if I could just sub in the $(\pi,\pi)$ throughout the calculation or if I have to find an equation for $(f\cdot g)(x,y)$ first and then find the value at $(\pi,\pi)$ from there, so I thought I would check here.