If $y=x^3log\, x $, then $\frac{d^2y}{dx^2}$ is equal to :

$x(5+6\, log x)$

The correct answer is Option (4) → $x(5+6\, log x)$

$y=x^3\log x$

$\frac{d}{dx}(uv)=u'v+uv'$

where $u=x^3$ & $v=\log x$

$\frac{du}{dx}=3x^2$

$\frac{dv}{dx}=\frac{1}{x}$

$\frac{dy}{dx}=x^3.\frac{1}{x}+3x^2.\log x$

$=x^2+3x^2\log x$

$=x^2(1+3\log x)$

Now, differentiate $\frac{dy}{dx}=x^2(1+3\log x)$ using product rule

$u=x^2$ & $v=(1+3\log x)$

Differentiate $u=x^2$

$\frac{du}{dx}=2x$

Differentiate $v=(1+3\log x)$

$\frac{dv}{dx}=\frac{3}{x}$

$\frac{d^2y}{dx^2}=2x(1+3\log x)+x^2.\frac{3}{x}$

$=2x+6x\log x+3x$

$=5x+6x\log x$

$=x(5+6\log x)$