Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 42

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


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


\end{aligned}
\end{equation}
$