Find the first derivative of $y=(x^4-1)\sqrt[3]{x^2-1}$ 
Find the first derivative of $$y=(x^4-1)\sqrt[3]{x^2-1}$$

We can write the function as $$y=(x^4-1)\left(x^2-1\right)^\frac13$$ For the derivative we have $$y'=4x^3\left(x^2-1\right)^\frac13+\dfrac13\left(x^2-1\right)^{-\frac23}2x(x^4-1)\\=4x^3\left(x^2-1\right)^\frac13+\dfrac23x\left(x^2-1\right)^{-\frac23}(x^2-1)(x^2+1)\\=\dfrac23x\left(x^2-1\right)^\frac13\left(6x^2+x^2+1\right)\\=\dfrac23x\left(x^2-1\right)^\frac13(7x^2+1)$$ The given answer is $$y'=\dfrac{2x(7x^4-6x^2-1)}{3\sqrt[3]{\left(x^2-1\right)^2}}$$ I don't see my mistake...
 A: The two formulas are equivalent.
$$\begin{align}
&\frac{2}{3}x\left(x^{2}-1\right)^{\frac{1}{3}}\left(7x^{2}+1\right)-\frac{2x\left(7x^{4}-6x^{2}-1\right)}{3\sqrt[3]{\left(x^{2}-1\right)^{2}}}
\\&=\frac{2x}{3\left(x^{2}-1\right)^{\frac{2}{3}}}\left(\left(7x^{2}+1\right)\left(x^{2}-1\right)^{\frac{1}{3}}\cdot\left(x^{2}-1\right)^{\frac{2}{3}}-\left(7x^{4}-6x^{2}-1\right)\right)
\\&=\frac{2x}{\ldots}\left(\left(7x^{2}+1\right)\left(x^{2}-1\right)-\left(7x^{4}-6x^{2}-1\right)\right)
\\&=0
\end{align}$$
It simply depends on whether you choose to bring together the sum given by the product rule in terms of $\left(x^{2}-1\right)^{\frac{1}{3}}$ or $\left(x^{2}-1\right)^{\frac{2}{3}}$. Any reasonable examiner should find both formulas correct, especially if they follow your correct working.
A: $$=\frac{2}{3}x\left(x^2-1\right)^{\frac{1}{3}}\left(7x^2+1\right)$$
$$=\frac{2}{3}x\left(x^2-1\right)^{\frac{1}{3}}\left(7x^2+1\right)\left(1\right)$$
$$=\frac{2}{3}x\left(x^2-1\right)^{\frac{1}{3}}\left(7x^2+1\right)\left(\frac{\left(x^2-1\right)^{\frac{2}{3}}}{\left(x^2-1\right)^{\frac{2}{3}}}\right)$$
$$=\frac{2}{3}x\left(x^2-1\right)^{\frac{1}{3}+\frac{2}{3}}\left(7x^2+1\right)\left(\frac{1}{\left(x^2-1\right)^{\frac{2}{3}}}\right)$$
$$=\frac{2}{3}x\left(x^2-1\right)\left(7x^2+1\right)\left(\frac{1}{^{\sqrt[3]{\left(x^2-1\right)^2}}}\right)$$
$$=\frac{2x}{3\sqrt[3]{\left(x^2-1\right)^2}}\left(x^2-1\right)\left(7x^2+1\right)$$
$$=\frac{2x\left(x^2-1\right)\left(7x^2+1\right)}{3\sqrt[3]{\left(x^2-1\right)^2}}$$
$$=\frac{2x\left(7x^4+x^2-7x^2-1\right)}{3\sqrt[3]{\left(x^2-1\right)^2}}$$
$$=\frac{2x\left(7x^4-6x^2-1\right)}{3\sqrt[3]{\left(x^2-1\right)^2}}$$
A: Too long for a comment
Make you life much easier using logarithmic differentiation
$$y=(x^4-1)\sqrt[3]{x^2-1} \implies \log(y)=\log(x^4-1)+\frac 13 \log(x^2-1)$$
$$\frac {y'}y=\frac {4x^3}{x^4-1}+\frac 13\frac {2x}{x^2-1}=\frac{2 x\left(7 x^2+1\right)}{3 \left(x^4-1\right)}$$
$$y'=\frac {y'}y \times y= ??? $$
