Implicit derivation to find $\partial x/\partial v$? I saw this question:
$$\begin{cases} x^2+y^2=u \\
x\sin y+y=v\end{cases}$$  
What is the $\partial x/\partial v$?   
I think it should be $1/\sin(y)$ because $\partial v/\partial x=\sin y$, but the answer is $$\frac{-y}{x^2\cos y+x-y\sin y}$$
But why?
 A: Be careful: you should consider $x$ and $y$ both to be functions of $u$ and $v$. Take the partial derivative with respect to $v$ of both equations in the system to get (don't forget the chain rule!):
$$\left\{ \begin{array}{rcl}
\displaystyle 2x\frac{\partial x}{\partial v}+2y\frac{\partial y}{\partial v}&=&0 \\[7pt]
\displaystyle \frac{\partial x}{\partial v}\sin y + x\cos y \frac{\partial y}{\partial v}+\frac{\partial y}{\partial v} &=& 1
\end{array}\right.$$
You can consider this as a system of two equations in the unknowns $\frac{\partial x}{\partial v}$ and $\frac{\partial y}{\partial v}$; solve for $\frac{\partial x}{\partial v}$.
A: Your idea is correct for total derivatives, but the equivalent of
$$
\frac{\mathrm dx}{\mathrm dv}=\frac1{\frac{\mathrm dv}{\mathrm dx}}
$$
is not
$$
\def\p#1#2{\frac{\partial#1}{\partial#2}}
\p xv=\frac1{\p vx}
$$
but the matrix equation
$$
\pmatrix{
\p xu&\p yu\\
\p xv&\p yv
}=
\pmatrix{
\p ux&\p vx\\
\p uy&\p vy
}^{-1}
$$
(which you can check by multiplying it out and applying the chain rule).
So you can indeed get the desired derivative without solving for $x$ and $y$, but you need to invert a matrix (or, equivalently, as in StackTD's answer, solve a system of linear equations).
