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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?

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    For instance, $\frac{\partial}{\partial x}u = f'(x-y)+g'(x+\frac{1}{3}y)$. This is just the chain rule for functions of one variable. Can you go on?2012-04-23
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    How do you find $\frac{\partial}{\partial y} u $ Using that method? or $ \frac{\partial^2}{\partial x \partial y} u $ ?2012-04-23
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    Well, $\frac{\partial}{\partial y}u=-f'(x-y)+\frac{1}{3}g'(x+\frac{1}{3}y)$, for example. The point is that $f$ and $g$ are functions of a single variable, so that you just need to differentiate as usual.2012-04-23
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    Ahhhhh right I've got you! Didn't quite click in my mind what you meant until you showed $ \frac{\partial}{\partial y} $ Thanks, I'll have another go at it, then mark you correct :D2012-04-23
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    The only problem I have still is with your $ ' $ notation? What does that mean? is $f'(x-y) = \frac{ \partial f(x-y)}{\partial y} $ or $ \frac{ \partial f(x-y)}{\partial x} $?2012-04-23
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    Because those are different things so when I do $ \frac{\partial^2u}{\partial x \partial y} u$ how do I notate that?2012-04-23
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    Ahhhhhhh I'm being stupid don't worry... It's neither is it? $f'(x-y) = \frac{\partial f(x-y)}{\partial s} $ ?2012-04-23

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