Higher Order Derivatives — Question 2

Question 2

Let $f(x) = x^2 e^{-x}$.

(a) Compute the first four derivatives of $f(x)$.
(b) Identify a pattern in the derivatives.
(c) Use the pattern to write a general formula for $f^{(n)}(x)$ for $n \geq 0$.

Question 2 - Solution

We are given $f(x) = x^2 e^{-x}$. Let’s compute derivatives step-by-step.

(a) First, compute the derivatives:

First derivative: $$f'(x) = \frac{d}{dx}(x^2 e^{-x}) = 2x e^{-x} + x^2 (-e^{-x}) = (2x - x^2) e^{-x}$$

Second derivative: $$\begin{gathered} f''(x) = \frac{d}{dx}[(2x - x^2)e^{-x}] = [2 - 2x] e^{-x} + (2x - x^2)(-e^{-x}) \\ = [2 - 2x - 2x + x^2] e^{-x} = (x^2 - 4x + 2) e^{-x} \end{gathered}$$

Third derivative: $$\begin{gathered} f^{(3)}(x) = \frac{d}{dx}[(x^2 - 4x + 2)e^{-x}] \\ = (2x - 4)e^{-x} + (x^2 - 4x + 2)(-e^{-x}) \\ = (-x^2 + 6x - 6)e^{-x} \end{gathered}$$

Fourth derivative: $$\begin{gathered} f^{(4)}(x) = \frac{d}{dx}[(-x^2 + 6x - 6)e^{-x}] \\ = (-2x + 6)e^{-x} + (-x^2 + 6x - 6)(-e^{-x}) \\ = (x^2 - 8x + 12)e^{-x} \end{gathered}$$

(b) Pattern:

Each derivative is a polynomial times $e^{-x}$. The polynomials alternate in sign and follow a recursive-like behavior: $$\begin{aligned} f(x) &= x^2 e^{-x} \\ f'(x) &= (2x - x^2)e^{-x} \\ f''(x) &= (x^2 - 4x + 2)e^{-x} \\ f^{(3)}(x) &= (-x^2 + 6x - 6)e^{-x} \\ f^{(4)}(x) &= (x^2 - 8x + 12)e^{-x} \end{aligned}$$

(c) General formula (structure):

Let: $$f^{(n)}(x) = P_n(x) e^{-x}$$ Where $P_n(x)$ is a polynomial of degree 2, alternating in sign: $$P_n(x) = (-1)^n(x^2 - 2nx + n(n-1))$$

Final boxed result: $$\boxed{f^{(n)}(x) = (-1)^n (x^2 - 2nx + n(n-1)) e^{-x}}$$