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Noliec
Group Title
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?)
 one year ago
 one year ago
Noliec Group Title
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?)
 one year ago
 one year ago

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Noliec Group TitleBest ResponseYou've already chosen the best response.2
@hartnn , @TuringTest , @AravindG Please help ;_;
 one year ago

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

AravindG Group TitleBest ResponseYou've already chosen the best response.0
It is right isnt it?
 one year ago

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

hartnn Group TitleBest ResponseYou've already chosen the best response.1
do u know componendo  dividendo ?
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
Take tanh x=y. Then apply log on both sides.
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
Solve for x.
 one year ago

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

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

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

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

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

hartnn Group TitleBest ResponseYou've already chosen the best response.1
solved without what ?
 one year ago

Noliec Group TitleBest ResponseYou've already chosen the best response.2
No shortcuts. :D
 one year ago

Noliec Group TitleBest ResponseYou've already chosen the best response.2
Algebraically
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
Great! Good work!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
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 = 1y need i go further ?
 one year ago

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

hartnn Group TitleBest ResponseYou've already chosen the best response.1
but did u get what i did ?
 one year ago

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

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

Noliec Group TitleBest ResponseYou've already chosen the best response.2
What happens after you apply ln?
 one year ago

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

Noliec Group TitleBest ResponseYou've already chosen the best response.2
Man I feel stupid. D:
 one year ago

Noliec Group TitleBest ResponseYou've already chosen the best response.2
Absolutely.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
any more doubts?
 one year ago

Noliec Group TitleBest ResponseYou've already chosen the best response.2
Nope, thanks a lot! :)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
welcome ^_^
 one year ago
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