Question

let $f(x)=x^{2}e^{x}$. $f(x)=\square$

Explanation:

Step1: Apply product rule

The product rule states that if \(y = u\cdot v\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = x^{2}\) and \(v = e^{x}\).
First, find \(u^\prime\): Using the power rule \((x^{n})^\prime=nx^{n - 1}\), so \(u^\prime=(x^{2})^\prime = 2x\).
Second, find \(v^\prime\): The derivative of \(e^{x}\) with respect to \(x\) is \(e^{x}\), so \(v^\prime=(e^{x})^\prime=e^{x}\).

Step2: Substitute into product rule formula

\(f^\prime(x)=u^\prime v+uv^\prime\)
Substitute \(u = x^{2}\), \(u^\prime = 2x\), \(v = e^{x}\), and \(v^\prime=e^{x}\) into the formula:
\(f^\prime(x)=2x\cdot e^{x}+x^{2}\cdot e^{x}\)
Factor out the common term \(e^{x}\):
\(f^\prime(x)=e^{x}(2x + x^{2})\)

Answer:

\(e^{x}(x^{2}+2x)\)