Why are partial derivatives of a harmonic function also harmonic? I've tried manipulating some expressions but still can't quite get my head around why the partial derivatives of $u(x,y)$, a harmonic function, are also harmonic.
 A: A harmonic function satisfies
$\sum_i\frac{\partial^2 f}{\partial x_i^2}=0$
Let's take a look at $g=\frac{\partial f}{\partial x_j}$ for some $j$. Then
\begin{align*}
\sum_i\frac{\partial^2 g}{\partial x_i^2}&=\sum_i\frac{\partial^3 f}{\partial x_i^2\partial x_j}\\
&=\sum_i\frac{\partial}{\partial x_j}\left(\frac{\partial^2 f}{\partial x_i^2}\right)\\
&=\frac{\partial}{\partial x_j}\sum_i\left(\frac{\partial^2 f}{\partial x_i^2}\right)\\
&=\frac{\partial}{\partial x_j}0\\
&=0
\end{align*}
and thus $g$ is harmonic.
This is of course assuming that the third partial derivatives of $f$ are well-defined.
Additional note: This proof requires that the order in which the partial derivatives are taken does not matter, i.e. $\frac{\partial^3 f}{\partial x_i^2\partial x_j} = \frac{\partial^3 f}{\partial x_j\partial x_i^2}$. I believe that this is not generally true (it does hold if the third partial derivative is continuous), so the proof only works for those functions where this is true.
A: Let $u$ be harmonic. If we investigate wether $u_x$ is harmonic we have to suppose that $u_x \in C^2$, hence we suppose $u \in C^3$. Let $v=u_x$
From $u_{xx}+u_{yy}=0$ we get by differentiatin w.r.t $x$:
$u_{xxx}+u_{yyx}=0$.
But this means: $v_{xx}+v_{yy}=0$ .
