MHB Partial Derivatives: Find $\frac{\partial^2{w}}{\partial{u}\partial{v}}$

[mathmari]

Hey!

Let $w=f(x, y)$ a two variable function and $x=u+v$, $y=u-v$.
Show that $$\frac{\partial^2{w}}{\partial{u}\partial{v}}=\frac{\partial^2{w}}{\partial{x^2}}-\frac{\partial^2{w}}{\partial{y^2}}$$

I have done the following:

We have $w(x(u,v), y(u, v))$.

From the chain rule we have: $$\frac{\partial{w}}{\partial{v}}=\frac{\partial{w}}{\partial{x}}\frac{\partial{x}}{\partial{v}}+\frac{\partial{w}}{\partial{y}}\frac{\partial{y}}{\partial{v}}=\frac{\partial{w}}{\partial{x}} \cdot 1+\frac{\partial{w}}{\partial{y}} \cdot (-1)=\frac{\partial{w}}{\partial{x}}-\frac{\partial{w}}{\partial{y}}$$

We apply again the chain rule, so:
$$\frac{\partial}{\partial{u}}\left (\frac{\partial{w}}{\partial{v}}\right )=\frac{\partial}{\partial{x}}\left (\frac{\partial{w}}{\partial{v}}\right ) \frac{\partial{x}}{\partial{u}}+\frac{\partial}{\partial{y}}\left (\frac{\partial{w}}{\partial{v}}\right )\frac{\partial{y}}{\partial{u}}=\frac{\partial}{\partial{x}}\left (\frac{\partial{w}}{\partial{x}}-\frac{\partial{w}}{\partial{y}}\right ) \cdot 1+\frac{\partial}{\partial{y}}\left (\frac{\partial{w}}{\partial{x}}-\frac{\partial{w}}{\partial{y}}\right ) \cdot 1\\ =\frac{\partial^2{w}}{\partial{x^2}}-\frac{\partial^2{w}}{\partial{x}\partial{y}}+\frac{\partial^2{w}}{\partial{x}\partial{y}}-\frac{\partial^2{w}}{\partial{y^2}}=\frac{\partial^2{w}}{\partial{x^2}}-\frac{\partial^2{w}}{\partial{y^2}}$$

Is it correct?? (Wondering)

Could I improve something at the formulation?? (Wondering)

 

[Opalg]

Looks good to me. (Yes)