Question

1. if $y = x^{2}(e^{x}-1)$, then $\frac{dy}{dx}=$

Explanation:

Step1: Apply the product rule

The product rule states that if \(y = u\cdot v\), then \(y^\prime=\frac{dy}{dx}=u^\prime v + uv^\prime\). Let \(u = x^{2}\) and \(v=e^{x}-1\). First, find \(u^\prime\) and \(v^\prime\).
\(u^\prime=\frac{d}{dx}(x^{2}) = 2x\) (using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\)), and \(v^\prime=\frac{d}{dx}(e^{x}-1)=e^{x}\) (since \(\frac{d}{dx}(e^{x})=e^{x}\) and \(\frac{d}{dx}(c)=0\) for a constant \(c\)).

Step2: Substitute into the product rule formula

\(\frac{dy}{dx}=u^\prime v+uv^\prime\)
\(\frac{dy}{dx}=2x(e^{x}-1)+x^{2}e^{x}\)
Expand the expression:
\(\frac{dy}{dx}=2xe^{x}-2x + x^{2}e^{x}\)
Factor out \(xe^{x}\):
\(\frac{dy}{dx}=xe^{x}(x + 2)-2x\)

Answer:

\(xe^{x}(x + 2)-2x\)