compute the differential dy for y = x³e²ˣ. your answer show include dx by typing dx. dy =

Explanation:

Step1: Apply product rule

The product rule states that if (y = u\cdot v), then (dy=(u'v + uv')dx). Let (u = x^{3}) and (v=e^{2x}). First, find (u') and (v'). (u'=\frac{d}{dx}(x^{3}) = 3x^{2}) (using the power rule (\frac{d}{dx}(x^{n})=nx^{n - 1})). (v'=\frac{d}{dx}(e^{2x})). Let (t = 2x), then (\frac{d}{dx}(e^{2x})=\frac{d(e^{t})}{dt}\cdot\frac{dt}{dx}). Since (\frac{d(e^{t})}{dt}=e^{t}) and (\frac{dt}{dx}=2), so (v' = 2e^{2x}).

Step2: Substitute into product rule formula

(dy=(u'v+uv')dx=(3x^{2}\cdot e^{2x}+x^{3}\cdot2e^{2x})dx). Factor out (x^{2}e^{2x}) from the expression inside the parentheses: (dy=x^{2}e^{2x}(3 + 2x)dx).

Answer:

(dy=x^{2}e^{2x}(3 + 2x)dx)