Calculus and Its Applications, Chapter 1, 1.8, Section 1.8, Problem 24

Determine the $f'(x)$ of the function $\displaystyle f(x) = \sqrt[3]{(x^2 - 1)^2}$
We have $f(x) =(x^2 - 1)^{\frac{2}{3}}$. So, by using Chain Rule

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


Then,

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