If $f(x) = x^3e^{-x}$, then the value of $f''(1)$ is equal to

$\frac{1}{e}$

The correct answer is Option (1) → $\frac{1}{e}$

$f(x)=x^{3}e^{-x}$

Differentiate:

$f'(x)=3x^{2}e^{-x}+x^{3}(-e^{-x})$

$f'(x)=e^{-x}(3x^{2}-x^{3})$

Differentiate again:

$f''(x)=-e^{-x}(3x^{2}-x^{3})+e^{-x}(6x-3x^{2})$

$f''(x)=e^{-x}\left[-(3x^{2}-x^{3})+(6x-3x^{2})\right]$

$f''(x)=e^{-x}(x^{3}-6x^{2}+6x)$

Evaluate at $x=1$:

$f''(1)=e^{-1}(1-6+6)$

$f''(1)=\frac{1}{e}$

Final answer: $\frac{1}{e}$