Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 41

Determine the derivative of the function $y = \sqrt{x+\sqrt{x}}$



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