Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 40

Differentiate $\displaystyle y = \arctan (\arcsin \sqrt{x})$


$
\begin{equation}
\begin{aligned}

y' =& \frac{d}{dx} [\arctan (\arcsin \sqrt{x})]
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y' =& \frac{1}{1 + (\arcsin \sqrt{x})^2} \cdot \frac{d}{dx} (\arcsin \sqrt{x})
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y' =& \frac{1}{1 + (\arcsin \sqrt{x})^2} \cdot \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \frac{d}{dx} (\sqrt{x})
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y' =&\frac{1}{1 + (\arcsin \sqrt{x})^2} \cdot \frac{1}{\sqrt{1 - x}} \frac{d}{dx} (x^{\frac{1}{2}})
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y' =& \frac{1}{\sqrt{1 - x} [1 + (\arcsin \sqrt{x})^2]} \cdot \frac{1}{2} x^{\frac{-1}{2}}
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y' =& \frac{1}{2 x^{\frac{1}{2}} \sqrt{1 - x} [1 + (\arcsin \sqrt{x})^2] }
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y' =& \frac{1}{2 x^{\frac{1}{2}} (1 - x)^{\frac{1}{2}} [1 + (\arcsin \sqrt{x})^2] }
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y' =& \frac{1}{2 (x - x^2)^{\frac{1}{2}} [1 + (\arcsin \sqrt{x})^2]}
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& \text{ or }
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y' =& \frac{1}{2 \sqrt{x - x^2} [1 + (\arcsin \sqrt{x})^2]}



\end{aligned}
\end{equation}
$