Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 26

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Determine the derivative of the function $y = \sqrt{x^2 - 1} \sec^{-1}(x)$ and simplify if possible.

If $y = \sqrt{x^2 - 1} \sec^{-1}(x)$, then by applying product rule and chain rule...

$ \begin{equation} \begin{aligned} y' &= (x^2 -1)^{\frac{1}{2}} \cdot \frac{d}{dx} \sec^{-1}(x) + \sec^{-1}(x) \cdot \frac{d}{dx} (x^2 -1)^{\frac{1}{2}}\\ \\ y' &= (x^2 -1)^{\frac{1}{2}} \cdot \left( \frac{1}{x\sqrt{x^2 - 1}} \right) + \sec^{-1}(x) \cdot \left[ \frac{1}{2}(x^2-1)^{-\frac{1}{2}} (2x) \right]\\ \\ y' &= \frac{1}{x} + \frac{x\sec^{-1}(x)}{\sqrt{x^2-1}} \end{aligned} \end{equation} $