Let $z=(x)^y$, what is $\frac{ \partial ^2z}{\partial x \partial y}$?

Question

Let $z=x^y$, what is $\displaystyle \frac{\partial ^2z}{\partial x \partial y}$ ?

My answer is
$ \displaystyle \frac{ \partial z}{ \partial y} = x^y \ln x$
$ \displaystyle \frac{ \partial ^2z}{\partial x \partial y} = \frac{x^y}{x} + \ln x(yx^{y-1})$

The choices are listed below but I don't seem to get the right answer. I am not sure if I have done it wrongly.

(A) 1
(B) $x/y$
(C) $(x)^y/x + y^2{(x)^y}^{-2}$
(D) $(x)^y/x$
(E) $y(x)^y$

Answer 1

Take logarithms:

$$\log z=y\log x$$

Then:

$$\frac{1}{z}\frac{\partial z}{\partial y}=\log x$$

From there you can compute the further derivatives.

Answer 2

You are correct. $$z=x^y$$

\begin{equation} \frac{\partial^2z}{\partial x\partial y}=\frac{\partial}{\partial x}x^y\ln x=\left(\frac{\partial}{\partial x}x^y\right)\ln x+x^y\frac{\partial}{\partial x}\ln x=yx^{y-1}\ln x+x^{y-1}=(y\ln x+1)x^{y-1} \end{equation}