Answer
$y'=\dfrac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}(1+\dfrac{1}{2\sqrt{x+\sqrt{x}}})(1+\dfrac{1}{2\sqrt{x}})$
Work Step by Step
$y=\sqrt{x+\sqrt{x+\sqrt{x}}}$
Let's rewrite the function like this:
$y=[x+(x+x^{1/2})^{1/2}]^{1/2}$
Differentiate using the chain rule:
$y'=\dfrac{1}{2}[x+(x+x^{1/2})^{1/2}]^{-1/2}[x+(x+x^{1/2})^{1/2}]'=...$
$...=\dfrac{1}{2}[x+(x+x^{1/2})^{1/2}]^{-1/2}[1+[(x+x^{1/2})^{1/2}]']=...$
Apply the chain rule one more time to find $[(x+x^{1/2})^{1/2}]'$:
$...=\dfrac{1}{2}[x+(x+x^{1/2})^{1/2}]^{-1/2}[1+[\dfrac{1}{2}(x+x^{1/2})^{-1/2}(x+x^{1/2})']]=...$
$...=\dfrac{1}{2}[x+(x+x^{1/2})^{1/2}]^{-1/2}[1+[\dfrac{1}{2}(x+x^{1/2})^{-1/2}(1+\dfrac{1}{2}x^{-1/2})]]$
Use algebra to simplify:
$...=\dfrac{1}{2[x+(x+x^{1/2})^{1/2}]^{1/2}}[1+[\dfrac{1}{2(x+x^{1/2})^{1/2}}][1+\dfrac{1}{2x^{1/2}}]]=$
$...=\dfrac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}(1+\dfrac{1}{2\sqrt{x+\sqrt{x}}})(1+\dfrac{1}{2\sqrt{x}})$
