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