# Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 33

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}$$x^{-1/3}, using the power rule \frac{d}{dx}[x^{4}-x^{3}-1]=4x^{3}-3x^{2}, using the power rule Now that we know that g'(x)=\frac{2}{3}$$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}}$

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