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...

Too long for a comment
Make you life much easier using logarithmic differentiation
$$y=(x^4-1)\sqrt[3]{x^2-1} \implies \log(y)=\log(x^4-1)+\frac 13 \log(x^2-1)$$
$$\frac {y'}y=\frac {4x^3}{x^4-1}+\frac 13\frac {2x}{x^2-1}=\frac{2 x\left(7 x^2+1\right)}{3 \left(x^4-1\right)}$$
$$y'=\frac {y'}y \times y= ??? $$
