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(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}} $

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(b) Please see proof.

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Okay. So that the reflective of Oxy gen x basest sort of is the stem. You start by writing it out, and you take impressive derivative on that. And so I use that. Cheryl, you're right. Why prime into himself, X However, there's that Stella you should hear Depends on how you pick your, uh, still men and their and the range when you're defining inverse function. Because in order to devising worse from shame must have want one function end up on If you defined the you need what one functioning depends on how you pick your moment And and this tension of oxy connects might be off by negative sign us is indicated here and I'll let you figure that out. I look a lot off should come true either pre cog bliss textbook or just to work out. Or you know how the sign changed when you pick a different moment to define owe a wrench to define the inverse function