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

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

By the product rule, with \(\frac{d}{dx}e^{-x} = -e^{-x}\):

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

Differentiating again by the product rule:

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

At \(x = 1\):

    \( \displaystyle f''(1) = e^{-1}(1 - 6 + 6) = \frac{1}{e} \)