Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 40

Differentiate $\displaystyle y = \frac{\sqrt{x} + 4}{\sqrt[3]{x} - 5}$
By applying Quotient Rule, we get

$\begin{equation} \begin{aligned} y' &= \frac{\left( x^{\frac{1}{3}} - 5 \right) \cdot \frac{d}{dx}\left( x^{\frac{1}{2}} + 4 \right) - \left( x^{\frac{1}{2}} + 4 \right) \cdot \frac{d}{dx} \left( x^{\frac{1}{3}} - 5 \right)}{\left( x^{\frac{1}{3}} - 5 \right)^2}\\ \\ y' &= \frac{\left( x^{\frac{1}{3}} - 5 \right) \left( \frac{1}{2} x^{-\frac{1}{2}} \right) - \left( x^{\frac{1}{2}} + 4 \right) \left(\frac{1}{3} x^{-\frac{2}{3}} \right)}{\left( x^{\frac{1}{3}} - 5 \right)^2}\\ \\ y' &= \frac{\frac{1}{2}x^{-\frac{1}{2} + \frac{1}{3}} - \frac{5}{2} x^{- \frac{1}{2}} - \frac{1}{3} x^{\frac{1}{2}-\frac{2}{3}} - \frac{4}{3} x^{-\frac{2}{3}} }{\left( x^{\frac{1}{3}} - 5 \right)^2}\\ \\ y' &= \frac{\frac{1}{2} x^{-\frac{1}{6}} - \frac{5}{2}x^{-\frac{1}{2}} - \frac{1}{3}x^{-\frac{1}{6}} - \frac{4}{3}x^{-\frac{2}{3}} }{\left( x^{\frac{1}{3}} - 5 \right)^2}\\ \\ y' &= \frac{\frac{1}{6}x^{-\frac{1}{6}} - \frac{5}{2}x^{-\frac{1}{2}} - \frac{4}{3} x^{-\frac{2}{3}} }{\left( x^{\frac{1}{3}} - 5 \right)^2}\\ \\ y' &= \frac{\frac{1}{6x^{\frac{1}{6}}} - \frac{5}{2x^{\frac{1}{2}}} - \frac{4}{3x^{\frac{2}{3}}}}{\left( x^{\frac{1}{3}} - 5 \right)^2}\\ \\ y' &= \frac{\frac{x^{\frac{7}{6}-\frac{1}{6}} - 5(3)\left( x^{\frac{7}{6}- \frac{1}{2}} \right) - 4(2) \left( x^{\frac{7}{6}- \frac{2}{3}} \right) }{6x^{\frac{7}{6}}} }{\left( x^{\frac{1}{3}} -5 \right)^2} && \text{Get the LCD } 6x^{\frac{7}{6}}\\ \\ y' &= \frac{x^{\frac{6}{6}} - 15x^{\frac{2}{3}} - 8x^{\frac{1}{2}} }{6x^{\frac{7}{6}}\left( x^{\frac{1}{3}} - 5 \right)^2 }\\ \\ y' &= \frac{x - 15x^{\frac{2}{3}} -8x^{\frac{1}{2}}}{6x^{\frac{7}{6}} \left( x^{\frac{1}{3}} - 5 \right)^2} \end{aligned} \end{equation}$