Answer
$y'=\dfrac{1}{2(1+\arcsin^ {2}\sqrt{x})\sqrt{x-x^{2}}}$
Work Step by Step
$y=\arctan(\arcsin\sqrt{x})$
Start the differentiation process by using the chain rule:
$y'=\dfrac{1}{1+(\arcsin\sqrt{x})^{2}}(\arcsin\sqrt{x})'=\dfrac{(\arcsin\sqrt{x})'}{1+\arcsin^{2}\sqrt{x}}=...$
Apply the chain rule one more time to evaluate the indicated derivative:
$...=\dfrac{\Big[\dfrac{1}{\sqrt{1-(\sqrt{x})^{2}}}\Big](\sqrt{x})'}{1+\arcsin^{2}\sqrt{x}}=\dfrac{\Big(\dfrac{1}{\sqrt{1-x}}\Big)(x^{1/2})}{1+\arcsin^{2}\sqrt{x}}=...$
Evaluate the derivative indicated and simplify:
$...=\dfrac{\Big(\dfrac{1}{\sqrt{1-x}}\Big)\Big(\dfrac{1}{2}\Big)x^{-1/2}}{1+\arcsin^{2}\sqrt{x}}=\dfrac{1}{2(\sqrt{x})(\sqrt{1-x})(1+\arcsin^{2}\sqrt{x})}=...$
$...=\dfrac{1}{2(1+\arcsin^ {2}\sqrt{x})\sqrt{x-x^{2}}}$
