Higher Order Derivatives — Question 7

Question 7 - Solution

We are given: $$f(x) = x^2 e^x$$

(a) Compute Derivatives:

We use the product rule: If $u = x^2$, $v = e^x$, then $f = uv$

$$f'(x) = (x^2)' e^x + x^2 (e^x) = 2x e^x + x^2 e^x = e^x(2x + x^2)$$

$$f''(x) = \frac{d}{dx}[e^x(2x + x^2)] = e^x(2 + 2x + 2x + x^2) = e^x(x^2 + 4x + 2)$$

$$f^{(3)}(x) = \frac{d}{dx}[e^x(x^2 + 4x + 2)] = e^x(x^2 + 4x + 2 + 2x + 4 + 0) = e^x(x^2 + 6x + 6)$$

(b) General Formula:

We notice a pattern in each derivative: $$f^{(n)}(x) = e^x \cdot P_n(x)$$ where $P_n(x)$ is a degree-2 polynomial (since the original function had $x^2$).

This suggests: $$\boxed{f^{(n)}(x) = e^x \cdot (x^2 + 2nx + n(n - 1))}$$

(This can be proven by induction.)

(c) Evaluate $f^{(3)}(0)$:

From (a): $$f^{(3)}(0) = e^0(0^2 + 6 \cdot 0 + 6) = 1 \cdot 6 = \boxed{6}$$