Prove $\frac{\partial u}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial v}{\partial x}\frac{\partial x}{\partial v}=1$ If $$u=f(x,y)$$
and
$$v=g(x,y)$$
are differentiable.
Prove $$\frac{\partial u}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial v}{\partial x}\frac{\partial x}{\partial v}=1$$
I knew that for one variable case 
$$\frac{du}{dx}=\frac{1}{\frac{dx}{du}}$$
 but I don't know how to do it in two variables. I also tried to differentiate both equations regarding to $x$, still no clue about how to solve it...Any help? Thank you~
 A: The answer by @W.mu is a poorly formulated, but correct.
Assume that the map
$$
(x,y) \mapsto (f(x,y),g(x,y))
$$
is invertible. That is there exist functions $d,e$ such that
$$
(u,v) \mapsto (d(u,v),e(u,v))
$$
is the inverse of the previous map, that is, giving $(x,y)$ as function of $(u,v)$. Thus, we have
$$
(x,y) \mapsto \Bigl(d\bigl(f(x,y),g(x,y)\bigr),e\bigl(f(x,y),g(x,y)\bigr)\Bigr) = (x,y).
$$
Then you can differentiate the first component of this composition with respect to $x$ to get
$$
  \frac{\partial d}{\partial u}\frac{\partial f}{\partial x}
 +\frac{\partial d}{\partial v}\frac{\partial g}{\partial x}
= \frac{\partial x}{\partial u}\frac{\partial u}{\partial x}
 +\frac{\partial x}{\partial v}\frac{\partial v}{\partial x} = 1.
$$
In the last step I have been sloppy in interpreting $x = x(u,v) = d(u,v,)$, etc. This is fairly common in mathematics writing, unfortunately, since otherwise you quickly run out of symbols to denote every new function, and the notation becomes very heavy. 
A: If we can suppose the transformation is invertible.
From IFT, we have:
$$x=x(u,v);y=y(u,v)$$
then we differentiate the first one to x:
$$1=x_uu_x+x_vv_x$$
Isn't it?
