- #1

- 673

- 29

Thus, ##\pm x=\cos^{-1}y## and ##ix=\cosh^{-1}y##.

So ##\cosh^{-1}y=\pm i\cos^{-1}y##.

Renaming the variable ##y##, we have ##\cosh^{-1}x=\pm i\cos^{-1}x##.

Next, we evaluate the derivative of ##\cosh^{-1}x## by converting it to ##\cos^{-1}x## using ##\cosh^{-1}x=i\cos^{-1}x##.

##\frac{d}{dx}\cosh^{-1}x=\frac{d}{dx}i\cos^{-1}x##

##=i\frac{-1}{\sqrt{1-x^2}}## --- (*)

##=\frac{-i}{i\sqrt{x^2-1}}## where I've used the rule ##\sqrt{-c}=i\sqrt{c}## for ##c>0##, since ##x>1##.

##=\frac{-1}{\sqrt{x^2-1}}##, which differs from the correct answer by a negative sign.

This means we would get the correct answer had we used ##\cosh^{-1}x=-i\cos^{-1}x##.

My question is how do we know when to add the negative sign.

I realized that step (*) is wrong because the formula for the derivative of ##\cos^{-1}## is true only for ##-1<x<1##. However, since we get the right answer in the end, it seems like there is a reason why we could still use the formula for values of ##x## outside ##-1<x<1##, and an explanation when we should use ##\cosh^{-1}x=-i\cos^{-1}x## instead of ##\cosh^{-1}x=i\cos^{-1}x##.