If $f(x) = x^3\log_e x$, Then $f''(e^2)$ is equal to

$17e^2$

The correct answer is Option (2) → $17e^2$

$f(x)=x^3\log_e x$

$f'(x)=3x^2\log_e x+x^3\cdot\frac{1}{x}$

$f'(x)=3x^2\log_e x+x^2$

$f''(x)=\frac{d}{dx}\left(3x^2\log_e x+x^2\right)$

$f''(x)=3\left(2x\log_e x+x^2\cdot\frac{1}{x}\right)+2x$

$f''(x)=3(2x\log_e x+x)+2x$

$f''(x)=6x\log_e x+3x+2x$

$f''(x)=6x\log_e x+5x$

At $x=e^2$

$\log_e(e^2)=2$

$f''(e^2)=6e^2\cdot2+5e^2$

$=12e^2+5e^2$

$=17e^2$

The value of $f''(e^2)$ is $17e^2$.