Answer
\[\frac{dy}{dx}=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}\left( 1+\frac{1}{2\sqrt{x+\sqrt{x}}}\left( 1+\frac{1}{2\sqrt{x}} \right) \right)\]
Work Step by Step
\[\begin{align}
& y=\sqrt{x+\sqrt{x+\sqrt{x}}} \\
& \text{Rewrite the function} \\
& y={{\left( x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right)}^{1/2}} \\
& \frac{dy}{dx}=\frac{d}{dx}\left[ {{\left( x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right)}^{1/2}} \right] \\
& \text{Differentiate by using the chain rule, then} \\
& \frac{dy}{dx}=\frac{1}{2}{{\left( x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right)}^{1/2-1}}\frac{d}{dx}\left[ x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right] \\
& \frac{dy}{dx}=\frac{1}{2}{{\left( x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right)}^{-1/2}}\left( 1+\frac{1}{2}{{\left( x+{{x}^{1/2}} \right)}^{-1/2}}\frac{d}{dx}\left[ x+{{x}^{1/2}} \right] \right) \\
& \text{Compute the derivative} \\
& \frac{dy}{dx}=\frac{1}{2}{{\left( x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right)}^{-1/2}}\left( 1+\frac{1}{2}{{\left( x+{{x}^{1/2}} \right)}^{-1/2}}\left( 1+\frac{1}{2\sqrt{x}} \right) \right) \\
& \text{Simplifying} \\
& \frac{dy}{dx}=\frac{1}{2{{\left( x+{{\left( x+{{x}^{1/2}} \right)}^{1/2}} \right)}^{1/2}}}\left( 1+\frac{1}{2{{\left( x+{{x}^{1/2}} \right)}^{1/2}}}\left( 1+\frac{1}{2\sqrt{x}} \right) \right) \\
& \text{Write the fractional exponents as radicals} \\
& \frac{dy}{dx}=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}\left( 1+\frac{1}{2\sqrt{x+\sqrt{x}}}\left( 1+\frac{1}{2\sqrt{x}} \right) \right) \\
\end{align}\]
