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Noliec
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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?)
 11 months ago
 11 months 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?)
 11 months ago
 11 months ago

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

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

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

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

AravindG Group TitleBest ResponseYou've already chosen the best response.0
Solve for x.
 11 months 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 ?
 11 months 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.
 11 months 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)
 11 months 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.
 11 months ago

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

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

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

Noliec Group TitleBest ResponseYou've already chosen the best response.2
Algebraically
 11 months ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
Great! Good work!
 11 months 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 ?
 11 months ago

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

hartnn Group TitleBest ResponseYou've already chosen the best response.1
but did u get what i did ?
 11 months 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
 11 months 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 (...)
 11 months ago

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

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

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

Noliec Group TitleBest ResponseYou've already chosen the best response.2
Absolutely.
 11 months ago

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

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

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