Calculus and Its Applications, Chapter 1, 1.7, Section 1.7, Problem 40

Differentiate $\displaystyle f(x) = \sqrt[3]{\frac{4-x^3}{x - x^2}}$.



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\begin{equation}
\begin{aligned}

f(x) =& \left( \frac{4-x^3}{x- x^2} \right)^{\frac{1}{3}}
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f'(x) =& \frac{1}{3} \left( \frac{4 - x^3}{x - x^2} \right)^{\frac{-2}{3}} \cdot \frac{d}{dx} \left( \frac{4 - x^3}{x - x^2} \right)
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f'(x) =& \frac{1}{3} \left( \frac{4 - x^3}{x - x^2} \right)^{\frac{-2}{3}} \left[ \frac{\displaystyle (x-x^2) \cdot \frac{d}{dx} (4-x^3) - (4-x^3) \cdot \frac{d}{dx} (x - x^2) }{(x - x^2)^2} \right]
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f'(x) =& \frac{1}{3} \left( \frac{4 - x^3}{x - x^2} \right)^{\frac{-2}{3}} \left[ \frac{(x - x^2) (-3x^2) - (4-x^3) (1 - 2x)}{(x - x^2)^2} \right]
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f'(x) =& \frac{1}{3} \left( \frac{4 - x^3}{x - x^2} \right)^{\frac{-2}{3}} \left[ \frac{-3x^3 + 3x^4 - 4 + 8x + x^3 - 2x^4}{(x - x^2)^2} \right]
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f'(x) =& \frac{1}{3} \left( \frac{4 - x^3}{x - x^2} \right)^{\frac{-2}{3}} \left[ \frac{x^4 - 2x^3 + 8x - 4}{(x - x^2)^2} \right]
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f'(x) =& \frac{x^4 - 2x^3 + 8x - 4}{3 (x - x^2)^{\frac{4}{3}} (4-x^3)^{\frac{2}{3}}}

\end{aligned}
\end{equation}
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