Answer
$$ y' =\frac{1}{8\sqrt{x}\sqrt{1+\sqrt{x}}\sqrt{1+\sqrt{1+\sqrt{x}}}}.$$
Work Step by Step
Since $ y=\sqrt{1+\sqrt{1+\sqrt{x}}}$, then by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$, the derivative $ y'$ is given by
$$ y'=\frac{(1+\sqrt{1+\sqrt{x}})'}{2\sqrt{1+\sqrt{1+\sqrt{x}}}}\\=
\frac{\frac{\frac{1}{2\sqrt{x}}}{2\sqrt{1+\sqrt{x}}}}{2\sqrt{1+\sqrt{1+\sqrt{x}}}}
\\=\frac{\frac{(1+\sqrt{x})'}{2\sqrt{1+\sqrt{x}}}}{2\sqrt{1+\sqrt{1+\sqrt{x}}}}\\ =\frac{1}{8\sqrt{x}\sqrt{1+\sqrt{x}}\sqrt{1+\sqrt{1+\sqrt{x}}}}.$$
