Verifying a formula for $\frac{\partial}{\partial u}w_x$, where $w = f(x,y)$, $x = u + v$, and $y = u - v$

Question

If I have $w = f(x,y)$, where $x = u + v$ and $y = u - v$, then why is the following true? $$\frac{\partial}{\partial u}w_x = \frac{\partial w_x}{\partial x}\cdot\frac{\partial x}{\partial u} + \frac{\partial w_x}{\partial y}\cdot\frac{\partial y}{\partial u}$$

Is my assumption that $\frac{\partial}{\partial u}w_x = \frac{\partial^2 w}{\partial u \partial x}$ incorrect since $w_x = \frac{\partial w}{\partial x}$?

Lastly,

How does that imply the following? $$\frac{\partial w_x}{\partial x}\cdot\frac{\partial x}{\partial u} + \frac{\partial w_x}{\partial y}\cdot\frac{\partial y}{\partial u} = w_{xx} + w_{xy}$$

Answer 1

If I have $w = f(x,y)$, where $x = u + v$ and $y = u - v$, then why is the following true? $$\frac{\partial}{\partial u}w_x = \frac{\partial w_x}{\partial x}\cdot\frac{\partial x}{\partial u} + \frac{\partial w_x}{\partial y}\cdot\frac{\partial y}{\partial u}$$

Is my assumption that $\frac{\partial}{\partial u}w_x = \frac{\partial^2 w}{\partial u \partial x}$ incorrect since $w_x = \frac{\partial w}{\partial x}$?

Not exactly. The notation of $\frac{\partial^2 w}{\partial u\partial x}$ should be avoided, since it is not a double derivative of independent variables ($x$ being dependent on $u$ and $v$).

Also, we should preferably use one style or the other, rather than mix notations, to avoid confusions.

$$\begin{align}\dfrac{\partial~~}{\partial u}\dfrac{\partial w}{\partial x}&=\dfrac{\partial^2 w}{\partial x\,^2}\cdot\dfrac{\partial x}{\partial u}+\dfrac{\partial^2 w}{\partial y\,\partial x}\cdot\dfrac{\partial y}{\partial u}\\[1ex] [w_x]_u &= w_{xx}x_u +w_{xy}y_u\end{align}$$

Lastly,

How does that imply the following? $$\frac{\partial w_x}{\partial x}\cdot\frac{\partial x}{\partial u} + \frac{\partial w_x}{\partial y}\cdot\frac{\partial y}{\partial u} = w_{xx} + w_{xy}$$

Well, because $x = u + v$ and $y = u - v$, therefore ...