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?)

- anonymous

- schrodinger

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- anonymous

@hartnn , @TuringTest , @AravindG
Please help ;_;

- anonymous

I basically get it, down to tanh(x)=(e^(2x)-1)/(e^(2x)+1)

- AravindG

It is right isnt it?

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## More answers

- anonymous

Indeed, but how do you go from here to the inverse of tanh(x)?

- hartnn

do u know componendo - dividendo ?

- AravindG

Take tanh x=y. Then apply log on both sides.

- AravindG

Solve for x.

- hartnn

if a/b = c/d
then
(a+b) /(a-b) = (c+d)/ (c-d)
did u know this ?

- anonymous

That's what I did, I didn't get anywhere on the log expression though. Hmm, not sure hartnn.

- hartnn

that rule is called componendo-dividendo
apply that for
y/1= (e^(2x)-1)/(e^(2x)+1)

- anonymous

I'd have to give a proof in order to utilize it though. D:
It's supposed to be solved without it.

- hartnn

that will isolate e^2x
and hence you can apply log

- hartnn

solved without what ?

- anonymous

No shortcuts. :D

- anonymous

Algebraically

- AravindG

Great! Good work!

- hartnn

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 ?

- hartnn

hint for next step
factor out e^2x from on left side

- hartnn

but did u get what i did ?

- hartnn

\(\huge \dfrac{1}{2}\ln (\dfrac{1+x}{1-x})\)
you should get this

- hartnn

-1-y / y-1 = (1+y)/ (1-y)
the 1/2 will come outside ln right ?
because
2x= ln (...)

- anonymous

What happens after you apply ln?

- hartnn

e^2x =[(1+y)/(1-y) ]
clear till here ??

- anonymous

Man I feel stupid. D:

- anonymous

Absolutely.

- hartnn

any more doubts?

- anonymous

Nope, thanks a lot! :)

- hartnn

welcome ^_^

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