# Partial Derivative Chain rule proof

Show that if $f$ is a function of the variables x and y (independent variables), and the latter are changed to independent variables u and v where $u = e^{y/x}$ and $x = x^2+y^2$, then

$x\frac{\partial{f}}{\partial{x}} + y\frac{\partial{f}}{\partial{y}} = 2v\frac{\partial{f}}{\partial{v}}$

I have no idea to start, I know how chain rule works for partial derivates when there the intermediate variables u and v are in terms of only one independent variable but I don't know what do to when it is in terms of two.

Can someone show me cause I have been stuck on this question for at least an hour.

Thank you so much!

Consider the function $f(x,y) = g(u,v) = g(e^{y/x}, x^2 + y^2)$.

Note also that $\frac{\partial u}{\partial x} = -\frac{ye^{y/x}}{x^2} = -\frac{yu}{x^2}$ and $\frac{\partial u}{\partial y} = \frac{e^{y/x}}{x} = \frac{u}{x}$ and $\frac{\partial v}{\partial x}= 2x$ and $\frac{\partial v}{\partial y}= 2y$.

Then $$\frac{\partial f}{\partial x} = \frac{\partial g}{\partial x} = \frac{\partial g}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial g}{\partial v}\frac{\partial v}{\partial x} = -\frac{yu}{x^2}\frac{\partial g}{\partial u} + 2x\frac{\partial g}{\partial v}$$

$$\frac{\partial f}{\partial y} = \frac{\partial g}{\partial y} = \frac{\partial g}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial g}{\partial v}\frac{\partial v}{\partial y} = \frac{u}{x}\frac{\partial g}{\partial u} + 2y\frac{\partial g}{\partial v}$$

So that

$$x\frac{\partial f}{\partial x} = -\frac{yu}{x}\frac{\partial g}{\partial u} + 2x^2\frac{\partial g}{\partial v}$$

$$y\frac{\partial f}{\partial y} = \frac{yu}{x}\frac{\partial g}{\partial u} + 2y^2\frac{\partial g}{\partial v}$$

Putting it together, knowing that $\frac{\partial f}{\partial v} = \frac{\partial g}{\partial v}$, we get the desired result:

$$x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = 2x^2\frac{\partial g}{\partial v} + 2y^2\frac{\partial g}{\partial v} = 2(x^2 + y^2)\frac{\partial g}{\partial v} = 2v\frac{\partial f}{\partial v}$$

• I dont quite understand why you are allowed to do: $\frac{\partial f}{\partial x} = \frac{\partial g}{\partial x} = \frac{\partial g}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial g}{\partial v}\frac{\partial v}{\partial x} = -\frac{yu}{x^2}\frac{\partial g}{\partial u} + 2x\frac{\partial g}{\partial v}$, dont u and v become the independent variables and x and y are intermediate values? – user2250537 Mar 22 '16 at 22:15
• Or does that depend on what you are trying to compute. – user2250537 Mar 22 '16 at 22:15
• Which part are you confused about? $u$ and $v$ depend on $x$ and $y$. – Adam Francey Mar 27 '16 at 19:33
• I got it thank you very much :) – user2250537 Mar 31 '16 at 20:42