Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 46

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


$
\begin{equation}
\begin{aligned}

y' =& \frac{d}{dx} (\sec h^{-1} \sqrt{1 - x^2})
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\\
y' =& \frac{1}{\sqrt{1 - x^2} \cdot \sqrt{1 - (\sqrt{1 - x^2})^2}} \cdot \frac{d}{dx} (\sqrt{1 - x^2})
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\\
y' =& \frac{1}{\sqrt{1 - x^2} \cdot \sqrt{1 - (1 - x^2)}} \cdot \frac{d}{dx} (1 - x^2)^{\frac{1}{2}}
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y' =& \frac{1}{\sqrt{1 - x^2} \cdot \sqrt{1 - 1 + x^2}} \cdot \frac{-1}{2} (1 - x^2)^{\frac{1}{2}} \cdot \frac{d}{dx} (1 - x^2)
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y' =& \frac{1}{\sqrt{1 - x^2} \cdot \sqrt{x^2}} \cdot \frac{1}{2} (1 - x^2)^{\frac{-1}{2}} \cdot (-2x)
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y' =& \frac{1}{(1 - x^2)^{\frac{1}{2}} \cancel{(x)}} \cdot \frac{- \cancel{2} \cancel{(x)}}{\cancel{2} (1 - x^2)^{\frac{1}{2}}}\
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y' =& \frac{-1}{1 - x^2}
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& \text{or}
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y' =& \frac{1}{x^2 - 1}

\end{aligned}
\end{equation}
$