Can someone explain how to calculate the third order partial derivative of $f.$

Question

$f(x,y)=\sin(xy).$ I calculated that $ \dfrac{\partial^2f}{\partial x\,\partial y}=\dfrac{\partial^2f}{\partial y\,\partial x}=\cos(xy)-xy\sin(xy)$.

I also calculated $$ \frac{\partial^3f}{\partial x^2\partial y}= -2y\sin(xy)-xy^2\cos(xy)$$ and $$\frac{\partial^3f}{\partial y^2\partial x}=-2x\sin(xy)-x^2y\cos(xy).$$

However I am not sure whether this is the correct method to calculate the third order partial derivative because $$ \frac{\partial^3f}{\partial x^2\partial y} \neq \frac{\partial^3f}{\partial y^2\partial x}.$$

Can someone explain how to calculate the third order partial derivative of $f.$

Answer 1

Define $g(x,y) = \dfrac{\partial^2 f}{\partial x \,\partial y}$.

Generally $\dfrac{\partial g}{\partial x} \neq \dfrac{\partial g}{\partial y}$.

By the way if you exchange the roles of $x$ and $y$, you get exactly the same as is to be expected from the form of $f$.

Answer 2

It will usually be the case that $\dfrac{\partial^3f}{\partial x^2 \, \partial y} \neq \dfrac{\partial^3f}{\partial y^2 \, \partial x}$. For this particular function, $f(x,y)=\sin(xy)$, the variables $x$ and $y$ play symmetrical roles. A consequence is that $\dfrac{\partial^3f}{\partial x^2 \, \partial y}$ will differ from $\dfrac{\partial^3f}{\partial y^2 \, \partial x}$ only in that $x$ and $y$ get interchanged.