# Multivariate Higher Order Derivatives

1. Dec 3, 2013

### Yagoda

1. The problem statement, all variables and given/known data
Let $h(u,v) = f(u+v, u-v)$. Show that $f_{xx} - f_{yy} = h_{uv}$ and $f_{xx} + f_{yy} = \frac12(h_{uu}+h_{vv})$.

2. Relevant equations

3. The attempt at a solution
I'm always confused on how to tackle these types of questions because there isn't an actual function to differentiate.
So I am assuming here that x = u+v and y = u-v and going from there. So I need to find the first partial with respect to x, which might be something like $f_x = \frac{\partial f}{\partial(u+v)}$ since I need to get it in terms of u and v, but this doesn't seem right. What sort of approach do I need to use on these questions?

Edit: What I've done now is write $h_{uv} = \frac{\partial}{\partial v}\frac{\partial h}{\partial u} = \frac{\partial}{\partial v} ( \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u}) =\frac{\partial}{\partial v} (\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y})$. I'm having trouble converting from u's and v's to x's and y's.

Last edited: Dec 3, 2013
2. Dec 3, 2013

### ShayanJ

$u=\frac{x+y}{2}$ and $v=\frac{x-y}{2}$ so $f(x,y)=h(\frac{x+y}{2},\frac{x-y}{2})$

So we have:

$f_x=h_u u_x+h_v v_x \Rightarrow f_x=\frac{1}{2}h_u+\frac{1}{2}h_v$

$f_{xx}=\frac{1}{2}(h_{uu}u_x+h_{uv}v_x+h_{vu}u_x+h_{vv}v_x)$

I think you can continue yourself.

3. Dec 3, 2013

### Dick

For thinking about these things it's sometimes good to think about the operators without worrying about the functions. Define, for example, $\partial_u(F)=\frac{\partial F}{\partial u}$. Then wouldn't it be true that $\partial_u=\partial_x+\partial_y$ and $\partial_v=\partial_x-\partial_y$ from the chain rule? It can save you a lot of texing and even spare some confusion. It's kind of the same as your underscore notation for partial derivatives.

Last edited: Dec 3, 2013