# Thread: A Multivariable chain rule question which is stomping me :(

1. ## A Multivariable chain rule question which is stomping me :(

The question goes like this:

Use the chain rule to verify the identity y * (∂g/∂x) + x * (∂g/∂y) = 0 if g(x,y) = f(x^2 - y^2, y^2 - x^2) and f is differentiable.

No answer key to help, so any support would be appreciated, thanks!

2. ## Re: A Multivariable chain rule question which is stomping me :(

Hey echodot.

What is the function g in this case?

3. ## Re: A Multivariable chain rule question which is stomping me :(

Originally Posted by echodot
Use the chain rule to verify the identity y * (∂g/∂x) + x * (∂g/∂y) = 0 if g(x,y) = f(x^2 - y^2, y^2 - x^2) and f is differentiable.
Denote $u=x^2-y^2,\;v=y^2-x^2.$ Then,

$\begin{Bmatrix} \dfrac{{\partial g}}{{\partial x}}=\dfrac{{\partial f}}{{\partial u}}\dfrac{{\partial u}}{{\partial x}}+\dfrac{{\partial f}}{{\partial v}}\dfrac{{\partial v}}{{\partial x}}=\dfrac{{\partial f}}{{\partial u}}(2x)+\dfrac{{\partial f}}{{\partial v}}(-2x)\\ \\\dfrac{{\partial g}}{{\partial y}}=\dfrac{{\partial f}}{{\partial u}}\dfrac{{\partial u}}{{\partial y}}+\dfrac{{\partial f}}{{\partial v}}\dfrac{{\partial v}}{{\partial y}}=\dfrac{{\partial f}}{{\partial u}}(-2y)+\dfrac{{\partial f}}{{\partial v}}(2y) \end{matrix}$

4. ## Re: A Multivariable chain rule question which is stomping me :(

@chiro, g is just some function.

@Fernando, thanks for the quick help, that makes complete sense, not sure why I hadn't done that earlier.