Answer
$u_{xy}=-\frac{2}{(x+2y)^{2}}$
and $u_{yx}=-\frac{2}{(x+2y)^{2}}$
Hence, $u_{xy}=u_{yx}$
Work Step by Step
Consider the function $u=ln(x+2y)$
Need to prove the conclusion of Clairaut’s Theorem holds, that is, $u_{xy}=u_{yx}$
In order to find this differentiate the function with respect to $x$ keeping $y$ constant.
$u_{x}=\frac{1}{(x+2y)}$
Differentiate the function with respect to $x$ keeping $y$ constant.
$u_{y}=\frac{2}{(x+2y)}$
Differentiate $u_{x}$ with respect to $y$ keeping $x$ constant.
$u_{xy}=\frac{∂}{∂y}[\frac{1}{(x+2y)}]$
$=-\frac{2}{(x+2y)^{2}}$
Thus, $u_{xy}=-\frac{2}{(x+2y)^{2}}$
Differentiate $u_{y}$ with respect to $x$ keeping $y$ constant.
$u_{yx}=\frac{∂}{∂x}[\frac{2}{(x+2y)}]$
$=-\frac{2}{(x+2y)^{2}}$
Therefore, $u_{xy}=-\frac{2}{(x+2y)^{2}}$
and $u_{yx}=-\frac{2}{(x+2y)^{2}}$
Hence, $u_{xy}=u_{yx}$
