Answer
(a)
$$
f(x)=\frac{3 x^{3}+6}{x^{2 / 3}}
$$
Then, by the quotient rule,
$$
\begin{aligned}
f^{\prime}(x)& =\frac{\left(x^{2 / 3}\right)\left(9 x^{2}\right)-\left(3 x^{3}+6\right)\left(\frac{2}{3} x^{-1 / 3}\right)}{\left(x^{2 / 3}\right)^{2}} \\
&=\frac{9 x^{8 / 3}-2 x^{8 / 3}-4 x^{-1 / 3}}{x^{4 / 3}} \\
&=\frac{7 x^{8 / 3}-\frac{4}{x^{43}}}{x^{4 / 3}} \\
&=\frac{7 x^{3}-4}{x^{5 / 3}}\\
&=7 x^{4 / 3}-4 x^{-5 / 3}
\end{aligned}
$$
(b)
$$
f(x)=f(x)=3 x^{7 / 3}+6 x^{-2 / 3}
$$
The derivative of a sum of functions is the sum of the derivatives. Then,we can find that,
$$
\begin{aligned}
f^{\prime}(x)&=3\left(\frac{7}{3} x^{4 / 3}\right)+6\left(-\frac{2}{3} x^{-5 / 3}\right) \\
&=7 x^{4 / 3}-4 x^{-5 / 3}
\end{aligned}
$$
(c)
The derivatives in (a) and (b) are equivalent.
Work Step by Step
(a)
$$
f(x)=\frac{3 x^{3}+6}{x^{2 / 3}}
$$
Then, by the quotient rule,
$$
\begin{aligned}
f^{\prime}(x)& =\frac{\left(x^{2 / 3}\right)\left(9 x^{2}\right)-\left(3 x^{3}+6\right)\left(\frac{2}{3} x^{-1 / 3}\right)}{\left(x^{2 / 3}\right)^{2}} \\
&=\frac{9 x^{8 / 3}-2 x^{8 / 3}-4 x^{-1 / 3}}{x^{4 / 3}} \\
&=\frac{7 x^{8 / 3}-\frac{4}{x^{43}}}{x^{4 / 3}} \\
&=\frac{7 x^{3}-4}{x^{5 / 3}}\\
&=7 x^{4 / 3}-4 x^{-5 / 3}
\end{aligned}
$$
(b)
$$
f(x)=f(x)=3 x^{7 / 3}+6 x^{-2 / 3}
$$
The derivative of a sum of functions is the sum of the derivatives. Then,we can find that,
$$
\begin{aligned}
f^{\prime}(x)&=3\left(\frac{7}{3} x^{4 / 3}\right)+6\left(-\frac{2}{3} x^{-5 / 3}\right) \\
&=7 x^{4 / 3}-4 x^{-5 / 3}
\end{aligned}
$$
(c)
The derivatives in (a) and (b) are equivalent.
