Calculus and Its Applications, Chapter 1, 1.8, Section 1.8, Problem 28

Determine the $y'$ of the function $\displaystyle y = (x^4 + x)^{\frac{2}{3}}$
By using Chain Rule,

$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} \left[ (x^4 + x)^{\frac{2}{3}} \right]\\
\\
&= \frac{2}{3} (x^4 + x)^{\frac{2}{3} - 1} \cdot \frac{d}{dx} (x^4 + x)\\
\\
&= \frac{2}{3} (x^4 + x)^{-\frac{1}{3}} (4x^3 + 1)\\
\\
&= \frac{2(4x^3 + 1)}{3(x^4 + x)^{\frac{1}{3}}}
\end{aligned}
\end{equation}
$


Then, by using Quotient Rule and Chain Rule

$
\begin{equation}
\begin{aligned}
y'' &= \frac{2}{3} \cdot \frac{d}{dx} \left[ \frac{4x^3 + 1}{(x^4 + x)^{\frac{1}{3}}} \right]\\
\\
&= \frac{2}{3} \left[ \frac{(x^4 + x)^{\frac{1}{3}} \cdot \frac{d}{dx} (4x^3 + 1) - (4x^3 + 1) \cdot \frac{d}{dx}
\left[ (x^4 + x)^{\frac{1}{3}} \right] }{\left[ (x^4 +x)^{\frac{1}{3}} \right]^2} \right]\\
\\
&= \frac{2}{3} \left[ \frac{(x^4 + x)^{\frac{1}{3}} (12x^2) - (4x^3 + 1) \left[ \frac{1}{2} (x^4 + x)^{-\frac{2}{3}} (4x^3 + 1) \right] }{\left[ (x^4 +x)^{\frac{1}{3}} \right]^2} \right]\\
\\
&= \frac{2}{3} \left[ \frac{(x^4 + x) (12x^2) - \frac{(4x^3 + 1)^2}{2(x^4 + x)^{\frac{2}{3}}} }{(x^4 + x)^{\frac{2}{3}}} \right]\\
\\
&= \frac{2}{3} \left[ \frac{2(12x^2) (x^4 + x)^{\frac{5}{3}} - (4x^3 + 1)^2 }{2(x^4 + x)^{\frac{2}{3}} (x^4 + x)^{\frac{2}{3}} }\right]\\
\\
&= \frac{1}{3} \left[ \frac{24x^2 (x^4 + x)^{\frac{5}{3}} - (4x^3 + 1)^2 }{(x^4 + x)^{\frac{4}{3}}} \right]
\end{aligned}
\end{equation}
$