Answer: (a) \(8xe^{4x}+2e^{4x}\). (b) \(\displaystyle \frac{xe^{4x}}4-\frac{e^{4x}}{16}+C\).

Answer: (a) \(8xe^{4x}+2e^{4x}\). (b) \(\displaystyle \frac{xe^{4x}}4-\frac{e^{4x}}{16}+C\).

(a) Use the product rule on \(y=2xe^{4x}\):

\(\displaystyle \frac{dy}{dx}=2e^{4x}+2x\cdot4e^{4x}\).

Therefore

\(\displaystyle \frac{dy}{dx}=2e^{4x}+8xe^{4x}\).

(b) From part (a),

\(\displaystyle \frac{d}{dx}(2xe^{4x})=2e^{4x}+8xe^{4x}\).

Rearrange this to isolate \(xe^{4x}\):

\(8xe^{4x}=\frac{d}{dx}(2xe^{4x})-2e^{4x}\).

Hence

\(\displaystyle \int xe^{4x}\,dx=\frac18(2xe^{4x})-\frac18\int2e^{4x}\,dx\).

So

\(\displaystyle \int xe^{4x}\,dx=\frac{xe^{4x}}4-\frac{e^{4x}}{16}+C\).