Question

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?)

Answer 1

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

Answer 2

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

Answer 3

It is right isnt it?

Answer 4

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

Answer 5

do u know componendo - dividendo ?

Answer 6

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

Answer 7

Solve for x.

Answer 8

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

Answer 9

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

Answer 10

that rule is called componendo-dividendo

apply that for
y/1= (e^(2x)-1)/(e^(2x)+1)

Answer 11

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

Answer 12

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

Answer 13

solved without what ?

Answer 14

No shortcuts. :D

Answer 15

Algebraically

Answer 16

Great! Good work!

Answer 17

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 ?

Answer 18

hint for next step

factor out e^2x from on left side

Answer 19

but did u get what i did ?

Answer 20

\(\huge \dfrac{1}{2}\ln (\dfrac{1+x}{1-x})\)

you should get this

Answer 21

-1-y / y-1 = (1+y)/ (1-y)

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

Answer 22

What happens after you apply ln?

Answer 23

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

Answer 24

Man I feel stupid. D:

Answer 25

Absolutely.

Answer 26

any more doubts?

Answer 27

Nope, thanks a lot! :)

Answer 28

welcome ^_^