I am trying to solve this problem here:
Consider the function of two variables:
$ u(x,y) = f(x−y) + g (x+ \frac{1}{3}y) $ , where $f(s)$ and $g(t)$ are each arbitrary functions of a single variable. Using the change of variables $s=x−y$, $t = x+\frac{1}{3}y$ use the chain rule to determine the first and second derivatives of u with respect to x and y in terms of derivatives of f and g. Hence, show that the second derivatives satisfy $u_{xx} = 2u_{xy} + 3u_{yy}$ where $u_{xx} = \frac{∂^2u}{∂x^2}$ etc.
I have tried this using the chain rules for partial derivatives ($\frac{\partial u}{\partial x} = \frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial t} \frac{\partial t}{\partial x}$)
So like this:
$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial s} + \frac{\partial u}{\partial t} \Rightarrow \frac{\partial^2u}{\partial x^2} = \frac{\partial(\frac{\partial u}{\partial s} + \frac{\partial u}{\partial t})}{\partial s} + \frac{\partial(\frac{\partial u}{\partial s} + \frac{\partial u}{\partial t})}{\partial t} = \frac{\partial^2u}{\partial s^2} + 2\frac{\partial^2u}{\partial s \partial t} + \frac{\partial^2u}{\partial t^2} $
I then did this for each double differential, however in doing this I get that:
$ 2u_{xy} + 3u_{yy} = \frac{\partial^2u}{\partial s^2} - \frac{10}{3}\frac{\partial^2u}{\partial s \partial t} + \frac{\partial^2u}{\partial t^2} $
Which is quite clearly wrong, is the way I am going about the question wrong? Or am I just making a silly mistake somewhere within my working?