I am trying to find $\cfrac{\partial^2 f}{\partial x\partial y}$ for the function: $$f=p(2x-y)+q(x-2y)$$ I have done double partial derivatives of functions $f(x,y)$ where $x(u,v)$ and $y(u,v)$ but this has kinda threw me. I am unsure of what question to ask as I feel like I am not understanding it properly. Hints on how to approach this would be awesome, thanks
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1Specify more. Are $p$ and $q$ functions or parameters? If they are functions, what Properties/regularity do they have? – Martingalo Feb 13 '14 at 20:58
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You simply have $$f=p(2x-y)+q(x-2y) \f_x=p'(2x-y)\cdot 2+q'(x-2y)\cdot 1$$ – Semsem Feb 13 '14 at 21:02
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Assuming $p$ and $q$ are differentiable functions then you just need to apply the chain rule:
$$\frac{\partial}{\partial x} f(x,y) = p'(2x-y)\cdot 2 + q'(x-2y)$$ where $p'$ and $q'$ denote the univariate derivatives.
Finally, we differentiable w.r.t. $y$ using the chain rule again: $$\frac{\partial^2}{\partial y \partial x} f(x,y) = -2p''(2x-y)-2q''(x-2y).$$
Here, I assume $p$ and $q$ are nice functions so $\frac{\partial^2}{\partial y \partial x} =\frac{\partial^2}{\partial x \partial y} $ as you can check.
Martingalo
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