Finding the partial derivatives of function

[chmate]

Homework Statement

If z=\frac{1}{x}[f(x-y)+g(x+y)], prove that \frac{\partial }{\partial x}(x^2\frac{\partial z}{\partial x})=x^2\frac{\partial^2 z}{\partial y^2}

Homework Equations

The Attempt at a Solution

I don't know how I'm supposed to find the partial derivative of z with respect to any of variables if the function f and g are not expressed.
Help me!

Thank you

 

[LCKurtz]

Homework Statement

If z=\frac{1}{x}[f(x-y)+g(x+y)], prove that \frac{\partial }{\partial x}(x^2\frac{\partial z}{\partial x})=x^2\frac{\partial^2 z}{\partial y^2}

Homework Equations

The Attempt at a Solution

I don't know how I'm supposed to find the partial derivative of z with respect to any of variables if the function f and g are not expressed.
Help me!

Thank you

Just leave the f and g in there, maybe they will cancel out. For example, if you want the partial of f(x-y) with respect to y it would be:$$
f_y(x-y) = -1\cdot f'(x-y)$$and so on.

 

[chmate]

Just leave the f and g in there, maybe they will cancel out. For example, if you want the partial of f(x-y) with respect to y it would be:$$
f_y(x-y) = -1\cdot f'(x-y)$$and so on.

Can you explain me why f_y(x-y) = -1\cdot f'(x-y)? I don't get it.
What is then f_x(x+y) = ?

 

[LCKurtz]

Can you explain me why f_y(x-y) = -1\cdot f'(x-y)? I don't get it.
What is then f_x(x+y) = ?

It is the chain rule. If you have $z=f(x-y)$ let $u = x-y$ so $z=f(u)$. Your chain rule for is$$
z_x = z_u u_x,\ z_y = f_u u_y$$and since f depends on only one variable u, you would write $f_u = f'(u)$ and you multiply by the $u_x$ or $u_y$ accordingly.