Define $f, g: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ as
\begin{align*}
f(x,y) := (x^2 + y^2, \sin(xy)) \\
g(x,y) := (xy, e^{y})
\end{align*}
Compute the derivative $D(g \circ f)$ in two ways: using the chain rule and by first writing out the map $g \circ f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ explicitly.
I need help for this question..
My idea:
First we will calculate the derivative $D(g \circ f)$ using the chain rule.
We have
\begin{align*}
Df(x,y) &= \begin{pmatrix} \frac{\partial f_1}{\partial x}(x^2 + y^2) & \frac{\partial f_1}{\partial y}(x^2 + y^2) \\ \frac{\partial f_2}{\partial x}(\sin(xy)) & \frac{\partial f_2}{\partial y}(\sin(xy)) \end{pmatrix}
= \begin{pmatrix} 2x & 2y \\ y\cos(xy) & x\cos(xy) \end{pmatrix} \\
\end{align*}
and
\begin{align*}
Dg(x,y) &= \begin{pmatrix} \frac{\partial g_1}{\partial x}(xy) & \frac{\partial g_1}{\partial y}(xy) \\ \frac{\partial g_2}{\partial x}(e^y) & \frac{\partial g_2}{\partial y}(e^y) \end{pmatrix}
= \begin{pmatrix} y & x \\ 0 & e^y \end{pmatrix}
\end{align*}
Now we can calculate $D(g \circ f)(x,y)$:
\begin{align*}
D(g \circ f)(x,y) &= Dg(f(x,y)) \circ Df(x,y) \\
& = Dg(x^2 + y^2, \sin(xy)) \circ Df(x,y) \\
&= \begin{pmatrix} \sin(xy) & x^2 + y^2 \\ 0 & e^{\sin(xy)} \end{pmatrix} \circ \begin{pmatrix} 2x & 2y \\ y\cos(xy) & x\cos(xy) \end{pmatrix} \\
&= \begin{pmatrix} 2x\sin(xy) + x^2y\cos(xy) + y^3 \cos(xy) & 2y\sin(xy) + x^3\cos(xy) + xy^2\cos(xy) \\ e^{\sin(xy)}y\cos(xy) & e^{\sin(xy)}x\cos(xy) \end{pmatrix}.
\end{align*}
Now we will calculate the derivative $D(g \circ f)(x,y)$ by writing out the map $g \circ f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$:
\begin{align*}
(g \circ f)(x,y) &= g(f(x,y)) \\
&= g(x^2 + y^2, \sin(xy)) \\
&= (x^2\sin(xy) + y^2\sin(xy), e^{\sin(xy)})
\end{align*}
We can find the derivative again:
\begin{align*}
D(g \circ f)(x,y) &= \begin{pmatrix} \frac{\partial}{\partial x} (x^2\sin(xy) + y^2\sin(xy)) & \frac{\partial}{\partial y} (x^2\sin(xy) + y^2\sin(xy)) \\ \frac{\partial}{\partial x} (e^{\sin(xy)}) & \frac{\partial}{\partial y} (e^{\sin(xy)})\end{pmatrix} \\
&= \begin{pmatrix} 2x\sin(xy) + x^2y\cos(xy) + y^3\cos(xy) & x^3\cos(xy) + 2y\sin(xy) + xy^2\cos(xy) \\ e^{\sin(xy)}y\cos(xy) & e^{\sin(xy)} x\cos(xy) \end{pmatrix}
\end{align*}
Both methods give the same derivative.
Best Answer
The mistake is when you construct the matrix $Dg(f(x,y))$, before multiplying it with $Df(x,y)$. You calculated the derivative $Dg$ correctly, but to use it in the chain rule correctly you have to evaluate it at $f(x,y)$, not at $(x,y)$.
Remember what a derivative is: it lets you estimate what small change in the output should result from a small change in the input. The input to the function $g$ is only a small change away $f(x,y)$ so to estimate the change in the output we need to use the derivative at $f(x,y)$. So you need to substitute $x^2 + y^2$ and $sin(xy)$ at the appropriate points.
One way to prevent this sort of error is to do what your original version of the question did that this version doesn't: use variables $u$ and $v$ so that $g(u,v) = (uv, e^v)$. This should help you avoid conflating the variables in the input to $f$ with the variables that are the input to $g$.