Answer
y'=$\frac{2(4x^{3}-3x^{2})}{(x^{4}-x^{3}-1)^{1/3}}$
Work Step by Step
The chain rule states that if y=g(h(x)), then y'=g'(h(x))h'(x). Since y=$(x^{4}-x^{3}-1)^{2/3}$, we can set g(x)=$x^{2/3}$ (outside function) and h(x)=$x^{4}$-$x^{3}$-1 (inside function), and use the chain rule to find y'. Therefore,
y'=g'($x^{4}$-$x^{3}$-1)h'(x)
We know that
$\frac{d}{dx}$[$x^{2/3}$]=$\frac{2}{3}$$x^{-1/3}$, using the power rule
$\frac{d}{dx}$[$x^{4}$-$x^{3}$-1]=4$x^{3}$-3$x^{2}$, using the power rule
Now that we know that g'(x)=$\frac{2}{3}$$x^{-1/3}$ and h'(x)=4$x^{3}$-3$x^{2}$, we can find y'.
y'=$\frac{2}{3}$$(x^{4}-x^{3}-1)^{-1/3}$(4$x^{3}$-3$x^{2}$)
=$\frac{2(4x^{3}-3x^{2})}{(x^{4}-x^{3}-1)^{1/3}}$
