# The double partial derivative

I have a differential of a function, $df(x,y)=y\sin(xy)\mathrm{dx}+x\sin(xy)\mathrm{dy}$.

How do I determine the double partial derivatives $f_{xy}, f_{xx}$ and $f_{yy}$?

I am fairly certain I have to use the chain rule, but I can't see how to apply because of the $y$ and $x$ in the front.

If it's true that I should use the chain rule, could you give me a hint?

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Note that $df(x,y)=f_{x}\mathrm{dx}+f_{y}\mathrm{dy}$, as $df(x,y)=y\sin(xy)\mathrm{dx}+x\sin(xy)\mathrm{dy}$, so we can get that $f_{x}=y\sin(xy)$ and $f_{y}=x\sin(xy)$, then $f_{xx}=y^{2}\cos(xy)$, $f_{yy}=x^{2}\cos(xy)$ and $f_{xy}=\sin(xy)+xy\cos(xy)$.
Thank you, I understand how you get $f_{xx}$ and $f_{yy}$, but why don't you get $f_{xy}=xy\cos(xy)$? – user37158 Oct 9 '12 at 16:50
@dexterity: Because $f_{xy}=(f_{x})_{y}$, it should derivate about $y$ to $y\sin(xy)$, note that $y\sin(xy)$ is a product of two functions of $y$, then you should use Lebniz rule. – Alfred Chern Oct 9 '12 at 16:56