Answer
$y'=\frac{1}{2}(\sqrt x)^{x}(1+lnx)$
Work Step by Step
Given: $y =(\sqrt x)^{x}$
Taking logarithmic on both sides of the function
$y =(\sqrt x)^{x}$
$lny=lnx^{\frac{x}{2}}=\frac{x}{2}lnx$
Take implicit differentiation with respect to $x$.
Apply product rule of differentiation.
$\frac{d}{dx}(lny)=\frac{d}{dx}(\frac{x}{2}lnx)$
$\frac{1}{y}\frac{d}{dx}(y)=\frac{x}{2}\frac{d}{dx}(lnx)+lnx\frac{d}{dx}(\frac{x}{2})$
$\frac{d}{dx}(y)=y[\frac{x}{2}\times(\frac{1}{x})+lnx\times\frac{1}{2}]$
Hence, $y'=\frac{1}{2}(\sqrt x)^{x}(1+lnx)$
