Question

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:

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