Calculus and Its Applications, Chapter 1, 1.8, Section 1.8, Problem 42

Find $y''$: For $\displaystyle y = x^{-3} + 2x^{\frac{1}{3}}$, find $f^{(5)} (x)$


$\begin{equation} \begin{aligned} f'(x) &= \frac{d}{dx} \left[ x^{-3} + 2x^{\frac{1}{3}} \right] = -3x^{-3-1} + 2 \cdot \frac{1}{3} x^{\frac{1}{3}-1} = -3x^{-4} + \frac{2}{3} x^{-\frac{2}{3}}\\ \\ f''(x) &= \frac{d}{dx} \left[ -3x^{-4} + \frac{2}{3} x^{-\frac{2}{3}} \right] = -3(-4)^{-4-1} + \frac{2}{3} \left( -\frac{2}{3}\right) x^{-\frac{2}{3}-1} = 12x^{-5} - \frac{4}{9}x^{-\frac{5}{3}}\\ \\ f'''(x) &= \frac{d}{dx} \left[ 12x^{-5} - \frac{4}{9}x^{-\frac{5}{3}} \right] = 12(-5)x^{-5-1} - \frac{4}{9} \left( - \frac{5}{3} \right)x^{-\frac{5}{3}-1} = -60x^{-6} + \frac{20}{27} x^{-\frac{8}{3}} \\ \\ f^{(4)}(x) &= \frac{d}{dx} \left[ -60x^{-6} + \frac{20}{27} x^{-\frac{8}{3}} \right] = -60 (-6)x^{-6-1} + \frac{20}{27}\left( - \frac{8}{3} \right)x^{-\frac{8}{3}-1} = 360x^{-7} - \frac{160}{81} x^{-\frac{11}{3}}\\ \\ f^{(5)}(x) &= \frac{d}{dx} \left[ 360x^{-7} - \frac{160}{81} x^{-\frac{11}{3}} \right] = 360 (-7)x^{-7-1} + \frac{160}{81}\left( - \frac{11}{3} \right)x^{-\frac{11}{3}-1} = 2520x^{-8} - \frac{1760}{243} x^{-\frac{14}{3}} \end{aligned} \end{equation}$