Calculate Derivative of a map 
Consider the maps from $R^2 \to R^2$ such that $F(u, v) = (e^{u + v}, e^{u - v})$ and $G(x, y) = (xy, x^2 - y^2)$. Calculate $D(F \circ G)(1, 1)$ by directly composing.

I got 
$F \circ G = (e^{x^2 +xy - y^2}, e^{y^2 + xy - x^2})$ 
But how do I get the derivative matrix?
 A: Remember that the derivative of a vector field is its Jacobian. If $F:\mathbb{R}^2\to\mathbb{R}^2$ is differentiable function with $F(x,y)=(f_1(x,y),f_2(x,y))$ where both $f_i$ are differentiable, then we have: $$D_{(x,y)}F = \left(
\begin{array}{cc}
 \frac{\partial}{\partial x}f_1(x,y) & \frac{\partial}{\partial y}f_1(x,y) \\
 \frac{\partial}{\partial x}f_2(x,y) & \frac{\partial}{\partial y}f_2(x,y) \\
\end{array}
\right)$$
For this problem, to calculate $D(F\circ G)$ you can either use the chain rule or calculate $F\circ G$ and then calculating the Jacobian directly like you are asking.

Solution
There are two ways to calculate the Jacobian for $F\circ G$.
First by calculating directly $D(F\circ G)$ which is:$$D(F\circ G) = D(e^{x^2 + xy - y^2},e^{y^2 + xy - x^2}) = \left(
\begin{array}{cc}
 e^{x^2+y x-y^2} (2 x+y) & e^{x^2+y x-y^2} (x-2 y) \\
 e^{-x^2+y x+y^2} (-2x+y) & e^{-x^2+y x+y^2} (x+2 y) \\
\end{array}
\right)$$
And we can also apply the chain rule by noticing that $D(F\circ G)=D(F(G))D(G) $ where $D(F(G))$ is $D(F)$ evaluated at $G$. We have:
$$D(F)= \left(
\begin{array}{cc}
 e^{x+y} & e^{x+y} \\
 e^{x-y} & -e^{x-y} \\
\end{array}
\right)\text{, } D(G) = \left(
\begin{array}{cc}
 y & x \\
 2 x & -2 y \\
\end{array}
\right)\text{ and }$$
$$D(F(G))= \left(
\begin{array}{cc}
 e^{x^2+y x-y^2} & e^{x^2+y x-y^2} \\
 e^{-x^2+y x+y^2} & -e^{-x^2+y x+y^2} \\
\end{array}
\right)$$
We multiply $D(F(G))D(G) $ to obtain:
$$\left(
\begin{array}{cc}
 e^{x^2+y x-y^2} & e^{x^2+y x-y^2} \\
 e^{-x^2+y x+y^2} & -e^{-x^2+y x+y^2} \\
\end{array}
\right)\left(
\begin{array}{cc}
 y & x \\
 2 x & -2 y \\
\end{array}
\right) = \left(
\begin{array}{cc}
 e^{x^2+y x-y^2} (y+2x) & e^{x^2+y x-y^2} (x-2 y) \\
 e^{-x^2+y x+y^2} (y-2 x) & e^{-x^2+y x+y^2} (x+2 y) \\
\end{array}
\right)$$
All that is left is to evalute $(x,y) = (1,1)$ which is: 
$$D_{(1,1)}(F\circ G) = \left(
\begin{array}{cc}
 3 e & -e \\
 -e & 3 e \\
\end{array}
\right)$$
