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

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

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

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