Answer
$y'=\frac{2+\ln x}{2\sqrt x}$
$y''=-\frac{\ln x}{4x\sqrt x}$
Work Step by Step
First derivative can be found by using product rule of differentiation.
Here, $y'=\frac{d}{dx}(\sqrt x\ln x)$
$=\sqrt x\frac{d}{dx}(\ln x)+\ln x\frac{d}{dx}{(\sqrt x)}$
$=\sqrt x.\frac{1}{x}+\ln x.[\frac{1}{2}(x^{-1/2})]$
Thus, $y'=\frac{2+\ln x}{2\sqrt x}$
After solving first derivative such as $y'=\frac{2+\ln x}{2\sqrt x}$, we will find second derivative with the help of quotient rule of differentiation.
$y''=\frac{1}{2}\frac{d}{dx}[\frac{2+\ln x}{2\sqrt x}]$
$=\frac{\sqrt x\frac{d}{dx}(2+\ln x)-(2+\ln x)\frac{d}{dx}\sqrt x}{(\sqrt x)^{2}}$
$=-\frac{1}{2}\frac{\ln x}{2x\sqrt x}$
$=-\frac{\ln x}{4x\sqrt x}$
Hence, $y''=-\frac{\ln x}{4x\sqrt x}$
