Answer
$$ y'= \frac{ 1}{4 \sqrt{1+x}\sqrt{\sqrt{1+x}+1}}.$$
Work Step by Step
Since $ y=\sqrt{\sqrt{1+x}+1}$, then by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$, the derivative $ y'$ is given by
$$ y'=\frac{(\sqrt{1+x}+1)'}{2\sqrt{\sqrt{1+x}+1}}
\\=\frac{\frac{1}{2\sqrt{1+x}}}{2\sqrt{\sqrt{1+x}+1}}=\frac{ 1}{4 \sqrt{1+x}\sqrt{\sqrt{1+x}+1}}.$$
