Inverse Hyperbolic Trigonometry as Logarithms: sech^-1(x)

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In this video I go over converting inverse hyperbolic trig functions to logarithms and prove that the function inverse hyperbolic secant or sech^-1(x) is equal to ln(1/x+sqrt(1-x^2)/x).

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I don't always deal with inverse hyperbolic functions but when I do I make sure to write them as logs ;)

mes
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In the begging the issue isn’t that you wouldn’t have a 1 to 1 function it’s that you wouldn’t have a function period. I think you meant that the issue with the original function is that it isn’t 1 to 1. Great video none the less.

DroughtBee
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so bad where are the details you are doing it fast !!
do it step by step

rizankobani