Answer: (a) \(x^2(3e^{3x})+2xe^{3x}\); (b)(i) \(2x(3x^2+4)^{-\frac23}\); (b)(ii) \(\frac12\left(\sqrt[3]{16}-\sqrt[3]{4}\right)\approx0.466\).
(a) Use the product rule on \(x^2e^{3x}\):
\(\frac{d}{dx}(x^2e^{3x})=x^2\frac{d}{dx}(e^{3x})+e^{3x}\frac{d}{dx}(x^2).\)
Therefore
\(\frac{d}{dx}(x^2e^{3x})=3x^2e^{3x}+2xe^{3x}.\)
(b)(i) Use the chain rule:
\(\frac{d}{dx}(3x^2+4)^{\frac13} =\frac13(3x^2+4)^{-\frac23}\cdot6x.\)
Hence
\(\frac{d}{dx}(3x^2+4)^{\frac13} =2x(3x^2+4)^{-\frac23}.\)
(b)(ii) From part (b)(i),
\(x(3x^2+4)^{-\frac23} =\frac12\frac{d}{dx}(3x^2+4)^{\frac13}.\)
So
\(\int_0^2 x(3x^2+4)^{-\frac23}\,dx =\left[\frac12(3x^2+4)^{\frac13}\right]_0^2.\)
Substitute the limits:
\(\frac12(16)^{\frac13}-\frac12(4)^{\frac13} =\frac12\left(\sqrt[3]{16}-\sqrt[3]{4}\right).\)
Numerically this is approximately
\(0.466.\)
