Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 28

Find $y'$ of $\displaystyle y = \frac{1}{\sqrt[3]{x+\sqrt{x}}}$



$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} \left( \frac{1}{\sqrt[3]{x+\sqrt{x}}} \right)\\
\\
y' &= \frac{\left( \sqrt[3]{x + \sqrt{x}} \right) \frac{d}{dx} (1) - (1) \frac{d}{dx} \left( \sqrt[3]{x + \sqrt{x}} \right) }{\left( \sqrt[3]{x + \sqrt{x}} \right)^2}\\
\\
y' &= \frac{(x + \sqrt{x})^{\frac{1}{3}} \frac{d}{dx} (1) - (1) \frac{d}{dx}(x + \sqrt{x})^{\frac{1}{3}} }{\left( \sqrt[3]{x + \sqrt{x}} \right)^2}\\
\\
y' &= \frac{(x + \sqrt{x})^{\frac{1}{3}}(0) - (1) \left(\frac{1}{3} \right)(x + \sqrt{x})^{\frac{-2}{3}}\frac{d}{dx} (x + \sqrt{x})}{(x + \sqrt{x})^{\frac{2}{3}} }\\
\\
y' &= \frac{-\frac{1}{3}(x + \sqrt{x})^{\frac{-2}{3}}\left( 1 + \frac{1}{2\sqrt{x}}\right) }{(x + \sqrt{x})^{\frac{2}{3}}}\\
\\
y' &= \frac{-(2 \sqrt{x}+1) }{3(2 \sqrt{x}) (x + \sqrt{x})^{\frac{2}{3}} (x+\sqrt{x})^{\frac{2}{3}} }\\
\\
y' &= \frac{-(2\sqrt{x}+1)}{6\sqrt{x}(x+\sqrt{x})^{\frac{4}{3}}}
\end{aligned}
\end{equation}
$