Simple partial differentiation I have a simple partial differentiation question here, given:
$u = x^2 - y^2$ and $v = x^2 -y$, find $x_u$ and $x_v$ in terms of $x$ and $y$.
What is the easiest way to go about this?
Thanks
 A: HINT: Think $x = x(u,v)$ and $y= y(u,v)$ and use implicit differentiation. 
First we consider the partial w.r.t. $u$
$$\left\{\begin{aligned}
\partial_u u &= \partial_u (x^2) - \partial_u (y^2),
\\
\partial_u v &= \partial_u (x^2) - \partial_u (y),
\end{aligned}\right.$$
this is:
$$\left\{\begin{aligned}
1 &= 2x \partial_u x - 2y\partial_u y,
\\
0 &= 2x \partial_u x  - \partial_u y.
\end{aligned}\right.$$
Now solve this equation for $\partial_u x$ and $\partial_u y$ in terms of $x$ and $y$:
$$
\partial_u x =\frac{1}{2x(1-2y)},\quad \partial_u y = \frac{1}{1-2y}.
$$
Now try to mimic above for the partial derivative w.r.t. $v$.
A: You can set up an inverse Jacobian matrix,
$$ J^{-1} \ = \ \left[ \begin{array}{cc}u_x&u_y\\v_x&v_y\end{array} \right]  $$
and then use whatever method you like to produce its inverse, for instance,
$$ J \ = \ \frac{1}{\det J^{-1}} \left[ \begin{array}{cc}v_y&-u_y\\-v_x&u_x\end{array} \right] \ = \ \left[ \begin{array}{cc}x_u&x_v\\y_u&y_v\end{array} \right] \ . $$
(This is not too much work when you only have two functions in two variables...)
For this problem, you'll have
$$ J^{-1} \ = \ \left[ \begin{array}{cc}2x&-2y\\2x&-1\end{array} \right] \ \Rightarrow \ \left[ \begin{array}{cc}x_u&x_v\\y_u&y_v\end{array} \right] \ = \  \frac{1}{4xy-2x} \left[ \begin{array}{cc}-1&2y\\-2x&2x\end{array} \right] \ . $$
[Note that this technique is a "shorthand" for the algebraically equivalent method that Shuhao Cao shows.]
