Prove $\frac{\partial u}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial v}{\partial x}\frac{\partial x}{\partial v}=1$

Question

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~

Answer 1

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?

Answer 2

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.