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Noliec
 2 years ago
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?)
Noliec
 2 years ago
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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Noliec
 2 years ago
Best ResponseYou've already chosen the best response.2@hartnn , @TuringTest , @AravindG Please help ;_;

Noliec
 2 years ago
Best ResponseYou've already chosen the best response.2I basically get it, down to tanh(x)=(e^(2x)1)/(e^(2x)+1)

Noliec
 2 years ago
Best ResponseYou've already chosen the best response.2Indeed, but how do you go from here to the inverse of tanh(x)?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1do u know componendo  dividendo ?

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.0Take tanh x=y. Then apply log on both sides.

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1if a/b = c/d then (a+b) /(ab) = (c+d)/ (cd) did u know this ?

Noliec
 2 years ago
Best ResponseYou've already chosen the best response.2That's what I did, I didn't get anywhere on the log expression though. Hmm, not sure hartnn.

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1that rule is called componendodividendo apply that for y/1= (e^(2x)1)/(e^(2x)+1)

Noliec
 2 years ago
Best ResponseYou've already chosen the best response.2I'd have to give a proof in order to utilize it though. D: It's supposed to be solved without it.

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1that will isolate e^2x and hence you can apply log

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1lol 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 = 1y need i go further ?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1hint for next step factor out e^2x from on left side

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1but did u get what i did ?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1\(\huge \dfrac{1}{2}\ln (\dfrac{1+x}{1x})\) you should get this

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.11y / y1 = (1+y)/ (1y) the 1/2 will come outside ln right ? because 2x= ln (...)

Noliec
 2 years ago
Best ResponseYou've already chosen the best response.2What happens after you apply ln?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1e^2x =[(1+y)/(1y) ] clear till here ??
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