I am trying to solve:
$x^{\sqrt{y}}=y^{\sqrt{x}}$
Here's my solution. Please correct me if there is an error and kindly explain why.
$\sqrt y\ln{x}=\sqrt x\ln{y}$
$\left[y^\frac{1}{2}\ln{x}\right]^\prime=\left[x^\frac{1}{2}\ln{y}\right]^\prime$
$\frac{1}{2}y^{-\frac{1}{2}}\ y^\prime\ln{x}+y^\frac{1}{2}\left(\frac{1}{x}\right)=\frac{1}{2}\ x^{-\frac{1}{2}}\ \ln{y}+x^\frac{1}{2}\left(\frac{1}{y}\right)y^\prime$
$\frac{\ln{x}}{2\sqrt y}\ y^\prime+\frac{\sqrt y}{x}=\frac{\ln{y}}{2\sqrt x}\ +\frac{\sqrt x}{y}y^\prime$
$\ \left(\frac{\ln{x}}{2\sqrt y}-\frac{\sqrt x}{y}\right)y^\prime=\frac{\ln{y}}{2\sqrt x}-\frac{\sqrt y}{x}\ $
$\left(\frac{y\ln{x}-2\sqrt{xy}}{2y\sqrt y}\right)y^\prime=\frac{x\ln{y}-2\sqrt{xy}}{2x\sqrt x}$
$y^\prime=\frac{y\sqrt y\left(x\ln{y}-2\sqrt{xy}\right)}{x\sqrt x\left(y\ln{x}-2\sqrt{xy}\right)}$
Thanks in advance.