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Numerade Educator



Problem 64 Medium Difficulty

(a) One way of defining $ \sec^{-1} x $ is to say that $ y= \sec^{-1} x \Leftrightarrow \sec y = x $ and $ 0 \le y < \pi/2 $ or $ \pi \le y < 3\pi/2. $ Show that, with this definition,
$ \frac {d}{dx} (\sec^{-1} x) = \frac {1}{x \sqrt {x^2 - 1}} $
(b) Another way of defining $ \sec^{-1} x \Leftrightarrow \sec y = x $ and $ 0 \le y \le \pi, y \not= \pi/2. $ Show that, with this definition,
$ \frac {d}{dx} (\sec{-1} x) = \frac {1}{\mid x \mid \sqrt {x^2 -1}} $


(a )Click to see
(b) Please see proof.


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Video Transcript

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