Answer
$y'=\frac{2+lnx}{2\sqrt x}$
and
$y''=-\frac{1}{2}\frac{lnx}{2x\sqrt x}$
Work Step by Step
First derivative can be found by using product rule of differentiation.
Here, $y'=\frac{d}{dx}(\sqrt xlnx)$
$=\sqrt x\frac{d}{dx}(lnx)+lnx\frac{d}{dx}{(\sqrt x)}$
$=\sqrt x.\frac{1}{x}+lnx.[\frac{1}{2}(x^{-1/2})]$
Thus, $y'=\frac{2+lnx}{2\sqrt x}$
After solving first derivative such as $y'=\frac{2+lnx}{2\sqrt x}$, we will find second derivative with the help of quotient rule of differentiation.
$y''=\frac{1}{2}\frac{d}{dx}[\frac{2+lnx}{2\sqrt x}]$
$=\frac{\sqrt x\frac{d}{dx}(2+lnx)-(2+lnx)\frac{d}{dx}\sqrt x}{(\sqrt x)^{2}}$
$=-\frac{1}{2}\frac{lnx}{2x\sqrt x}$
Hence, $y''=-\frac{1}{2}\frac{lnx}{2x\sqrt x}$
