Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 44

Determine the derivative of $\displaystyle y = x \tan h^{-1} x + \ln \sqrt{1 - x^2}$. Simplify where possible.


$
\begin{equation}
\begin{aligned}

y' =& \frac{d}{dx} (x \tan h^{-1} x) + \frac{d}{dx} (\ln \sqrt{1 - x^2})
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y' =& \left[ x \cdot \frac{d}{dx} (\tan h^{-1} x) + \tan h^{-1} x \cdot \frac{d}{dx} (x) \right] + \frac{1}{\sqrt{1 - x^2}} \cdot \frac{d}{dx} (\sqrt{1 - x^2})
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y' =& x \cdot \frac{1}{1 - x^2} + \tan h^{-1} x + \frac{1}{(1 - x^2)^{\frac{1}{2}}} \cdot \frac{d}{dx} (1 - x^2)^{\frac{1}{2}}
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y' =& \frac{x}{1 - x^2} + \tan h^{-1} x + \frac{1}{(1 - x^2)^{\frac{1}{2}}} \cdot \frac{1}{2} (1 - x^2)^{\frac{-1}{2}} \cdot \frac{d}{dx} (1 - x^2)
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y' =& \frac{x}{1 - x^2} + \tan h^{-1} x + \frac{1}{(1 - x^2)^{\frac{1}{2}}} \cdot \frac{1}{\cancel{2}} (1 - x^2)^{\frac{-1}{2}} \cdot -\cancel{2} x
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y' =& \frac{x}{1 - x^2} + \tan h^{-1} x - \frac{x}{(1 - x^2)^{\frac{1}{2}} (1 - x^2)^{\frac{1}{2}}}
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y' =& \cancel{ \frac{x}{1 - x^2} } + \tan h^{-1} x - \cancel{ \frac{x}{1 - x^2} }
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y' =& \tan h^{-1} x

\end{aligned}
\end{equation}
$