1) We need to differentiate the following functions:
[tex]\begin{gathered} a)\:f(x)=x\sqrt[3]{1+x^2}\:\:\:\:Use\:the\:product\:rule \\ \\ \\ \frac{d}{dx}\left(x\right)\sqrt[3]{1+x^2}+\frac{d}{dx}\left(\sqrt[3]{1+x^2}\right)x \\ \\ \\ 1\cdot \sqrt[3]{1+x^2}+\frac{2x}{3\left(1+x^2\right)^{\frac{2}{3}}}x \\ \\ \sqrt[3]{1+x^2}+\frac{2x^2}{3\left(x^2+1\right)^{\frac{2}{3}}} \\ \\ f^{\prime}(x)=\sqrt[3]{1+x^2}+\frac{2x^2}{3\left(1+x^2\right)^{\frac{2}{3}}} \end{gathered}[/tex]Note that we had to use some properties like the Product Rule, and the Chain Rule.
b) We can start out by applying the Quotient Rule:
[tex]\begin{gathered} g(x)=\frac{(x^2-1)^3}{(2x+1)} \\ \\ f^{\prime}(x)=\frac{\frac{d}{dx}\left(\left(x^2-1\right)^3\right)\left(2x+1\right)-\frac{d}{dx}\left(2x+1\right)\left(x^2-1\right)^3}{\left(2x+1\right)^2} \\ \\ Differentiating\:each\:part\:of\:that\:quotient: \\ \\ ------- \\ \frac{d}{dx}\left(\left(x^2-1\right)^3\right)=3\left(x^2-1\right)^2\frac{d}{dx}\left(x^2-1\right)=6x\left(x^2-1\right)^2 \\ \\ \frac{d}{dx}\left(x^2-1\right)=\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(1\right)=2x \\ \\ \frac{d}{dx}\left(x^2\right)=2x \\ \\ \frac{d}{dx}\left(1\right)=0 \\ \\ \frac{d}{dx}\left(2x+1\right)=2 \\ \\ Writing\:all\:that\:together: \\ \\ f^{\prime}(x)=\frac{6x\left(x^2-1\right)^2\left(2x+1\right)-2\left(x^2-1\right)^3}{\left(2x+1\right)^2} \\ \end{gathered}[/tex]Thus, these are the answers.
