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

Given $f(x) = x^3 e^{-x}$, find $f''(1)$.

Since $f(x) = x^3 \cdot e^{-x}$, use the Product Rule with:

• $u = x^3$ → $u' = 3x^2$
• $v = e^{-x}$ → $v' = -e^{-x}$

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

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

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

Differentiate $f'(x) = e^{-x}(3x^2 - x^3)$ using the Product Rule with:

• $u = e^{-x}$ → $u' = -e^{-x}$
• $v = 3x^2 - x^3$ → $v' = 6x - 3x^2$

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

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

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

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

Evaluate at $x = 1$:

$f''(1) = e^{-1}[1^3 - 6(1)^2 + 6(1)]$

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

$f''(1) = e^{-1}[1]$

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

Therefore, $f''(1) = \dfrac{1}{e}$