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# Osborn's rule and proving watch

1. i am stuck on this question:

show that:

tanh^-1(x)=1/2(ln(1+x/1-x))

what i done was..

x=tanhy

dx/dy=sech^2y and so therefore dy/dx=1/sech^2y

then using the identity: sech^y-1=tanh^2y

i got: dy/dx=sech^2y-tanh^2y/tanh^2y+1

but now i am stuck... any help please
2. It's an algebra problem - you shouldn't be differentiating. Instead:

x = tanh y

.

Then, solve for y.

You may find it useful to write Y = e^2y and solve for Y, then take logs to find y.
3. Why are you differentiating?

Express x = tanh y in terms of exponentials and change the subject.
4. (Original post by DFranklin)
It's an algebra problem - you shouldn't be differentiating. Instead:

x = tanh y

.

Then, solve for y.

You may find it useful to write Y = e^2y and solve for Y, then take logs to find y.
ok so make x equal to the exponentials etc and then get Y and take log to work out what y equals to and i should end up with that answer... hmm i shall attempt it thank you
5. did you mean by the way turn it into this then work out:

x=y-1/y+1 ?
6. (Original post by sahil112)
did you mean by the way turn it into this then work out:

x=y-1/y+1 ?
You are going to confuse yourself by using y to mean two different things. Anyway multiply both sides by y + 1, rearrange and factorise.
7. ok i have ended up with y=-1-x/x-1

i think i should sub x back into it now but dont no if that will make it very complicated?
8. (Original post by sahil112)
ok i have ended up with y=-1-x/x-1

i think i should sub x back into it now but dont no if that will make it very complicated?
Y= x+1/1-x

But Y=e^2y.

So e^2y= x+1/1-x

I guess you can do the rest?

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Updated: November 27, 2011
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