Answer
$$ y' =\frac{1+\frac{1+\frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}}}{2\sqrt{x+\sqrt{x+\sqrt{x}}}} .$$
Work Step by Step
Since $ y=\sqrt{x+\sqrt{x+\sqrt{x}}}=(x+\sqrt{x+\sqrt{x}})^{1/2}$, the derivative $ y'$, by using the chain rule $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$, is given by
$$ y'=\frac{1}{2}(x+\sqrt{x+\sqrt{x}})^{-1/2}(x+\sqrt{x+\sqrt{x}})'
\\=\frac{1+\frac{(x+\sqrt{x})'}{2\sqrt{x+\sqrt{x}}}}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}
\\=\frac{1+\frac{1+\frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}}}{2\sqrt{x+\sqrt{x+\sqrt{x}}}} .$$
