Application of the Inverse Function Theorem to polar co-ordinate transformation $U=\{(u,v)\in {R}^2:u>0\}$ 
Define a function $F:U \rightarrow R^2$ by $F(u,v)= (u\cos(v),u\sin(v))= (x,y)$
a) Show $F$ is an open mapping on $U$. [I've done this.]
b) Calculate $\Large \frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}$.
Partial result: I think I can write $u$ in terms of $x$ and $y$ by doing the following: $$x^2+y^2=u^2(\sin^2(v)+\cos^2(v))$$ Thus we have $u=\sqrt{x^2+y^2}$. Then I could differentiate with respect to $x$ of course. 
However, I think there is a way to compute these partials directly from the derivative of $F$ using the Inverse Function Theorem, but I don't know how.
 A: By the Inverse Function Theorem, the matrices
$$
\begin{bmatrix}
\dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \\
\dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y}
\end{bmatrix}
\qquad\text{and}\qquad
\begin{bmatrix}
\dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \\
\dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v}
\end{bmatrix}
$$
should be inverses.  The second matrix is easy to compute:
$$
\begin{bmatrix}
\dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \\
\dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v}
\end{bmatrix}
\;\;=\;\;
\begin{bmatrix}
\cos v & -u \sin v \\
\sin v & u\cos v
\end{bmatrix}
$$
Inverting yields:
$$
\begin{bmatrix}
\dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \\
\dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y}
\end{bmatrix}
\;\;=\;\;
\frac{1}{u}
\begin{bmatrix}
u\cos v & u \sin v \\
-\sin v & \cos v
\end{bmatrix}
$$
If we want these partial derivatives in terms of $x$ and $y$, we can substitute $\cos v = x/u$ and $\sin v = y/u$ to get
$$
\begin{bmatrix}
\dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \\
\dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y}
\end{bmatrix}
\;\;=\;\;
\begin{bmatrix}
x/u & y/u \\
-y/u^2 & x/u^2
\end{bmatrix}
$$
where $u=\sqrt{x^2+y^2}$.
