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I found a question:

Suppose that $z= e^{x+y}$. Show that the result of differentiation $z$, $m$ times with respect to $x$ and $n$ times with respect to $y$ is

$$\frac{\partial^{m+n}z}{\partial x^m\,\partial y^n} = e^{x+y}$$

When I think about it, it is right because any differentiation either with respect to $x$ or $y$ is still $e^{x+y}$, and through the chain rule of $x+y$ , it will always become 1. So, it will be $e^{x+y}$ times Q $1$’s every time you differentiate it $Q$ times. Now, how does one show it is correct.

Behrouz Maleki
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    you could write $z(x,y)=\exp(x)\exp(y)$. Now, the result is really trivial, since you can split of the partial derivatives...on the two factors and use the one dimensional result. – Alex Jul 10 '16 at 19:23
  • What Alex said plus induction on either variables, or both, if you wish. – Git Gud Jul 10 '16 at 19:27

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