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Does anyone know where you could find the deduction of the inverse of tanh(x) starting from the standard hyperbolic trigonometry definitions? (Or does anyone possibly know how to deduce it from them?)

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@hartnn , @TuringTest , @AravindG Please help ;_;
I basically get it, down to tanh(x)=(e^(2x)-1)/(e^(2x)+1)
It is right isnt it?

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Other answers:

Indeed, but how do you go from here to the inverse of tanh(x)?
do u know componendo - dividendo ?
Take tanh x=y. Then apply log on both sides.
Solve for x.
if a/b = c/d then (a+b) /(a-b) = (c+d)/ (c-d) did u know this ?
That's what I did, I didn't get anywhere on the log expression though. Hmm, not sure hartnn.
that rule is called componendo-dividendo apply that for y/1= (e^(2x)-1)/(e^(2x)+1)
I'd have to give a proof in order to utilize it though. D: It's supposed to be solved without it.
that will isolate e^2x and hence you can apply log
solved without what ?
No shortcuts. :D
Great! Good work!
lol ok, don't apply that rule (bdw, that was algebraic property) y= (e^(2x)-1)/(e^(2x)+1) y (e^2x +1) = e^2x -1 y e^2x + y = e^2x -1 y e^2x -e^2x = -1-y need i go further ?
hint for next step factor out e^2x from on left side
but did u get what i did ?
\(\huge \dfrac{1}{2}\ln (\dfrac{1+x}{1-x})\) you should get this
-1-y / y-1 = (1+y)/ (1-y) the 1/2 will come outside ln right ? because 2x= ln (...)
What happens after you apply ln?
e^2x =[(1+y)/(1-y) ] clear till here ??
Man I feel stupid. D:
any more doubts?
Nope, thanks a lot! :)
welcome ^_^

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