1. ## hyperbolic differentiation

Q (i)prove that d/dx(artanhx)=1/(1-x^2)

this is just artanhx=y, tanhy=x then diff.

can't do simplify end of (ii) Q by using partial ractions and integrating deduce from this(ans to (i)) the logarithmic form of artanhx,,,i know the log form of artanhx is 1/2ln(1+x/1-x)???????

i did 1/(1-x^2) = (int)-1/(-2x-2) -(int)1/(2x-2)

giving 1/2ln|-2x-2| -ln|2x-2| then i get to 1/2ln|(-2x-2)/(2x-2)| not sure where to go from here i think its easy but im being ignorant,,,the log form of artanhx is 1/2ln(1+x/1-x)???????pls help!!!!

2. Let's start by proving that $\displaystyle tanh^{-1}(x)=\frac{1}{2}ln(\frac{1+x}{1-x})$

Let $\displaystyle y=tanh^{-1}(x), \;\ x=tanh(y)=\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}=\frac{e^{2y}-1}{e^{2y}+1}$

$\displaystyle xe^{2y}+x=e^{2y}-1$

$\displaystyle e^{2y}(x-1)=-x-1,$

$\displaystyle e^{2y}=\frac{1+x}{1-x}, \;\ 2y=ln(\frac{1+x}{1-x}), \;\ y=\frac{1}{2}ln(\frac{x+1}{x-1})$

Now, $\displaystyle \frac{d}{dx}[tanh^{-1}(x)]=\frac{d}{dx}[\frac{1}{2}(ln(1+x)-ln(1-x))]=\frac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x}\right)=\frac{1}{1-x^{2}}$

3. ## cheers

i used (-x-1)(x-1) for the 'base' of my partial fractions,,, shudda used (x+1)(-x+1) cheers fella!

4. ## :o

oh no wait when i use either the partial expantion of (x+1)(-x+1) or the base of (-x-1)(x-1) i still cant expand to get this,,
with (x+1)(-x+1) i get 1/2ln(2x+2)+l(x-1)

and with (-x-1)(x-1) i get 1/2ln(-2x-2)-1/2ln(2x-2) u know where to go from here??

5. They are the same:

$\displaystyle \frac{-x-1}{x-1}=\frac{1+x}{1-x}$