1. ## Proof of derivative

I'm stuck on this question. How would I go about doing it? Thanks in advance.

Q: Prove that the derivative of $\displaystyle \mathrm{artanh} x$, $\displaystyle -1 < x < 1$, is $\displaystyle \frac{1}{1-x^2}$.

2. Start by showing that $\displaystyle \text{arctanh }x=\frac{1}{2}\ln \frac{1+x}{1-x}.$ After that contemplate its derivative.

3. Hello,

You can do this way :

$\displaystyle y=\tanh(x)=\frac{\sinh(x)}{\cosh(x)}=\frac{e^x-e^{-x}}{2} \cdot \frac{2}{e^x+e^{-x}}= \boxed{\frac{e^x-e^{-x}}{e^x+e^{-x}}}$

$\displaystyle x=\tanh^{-1}(y)$

So we have to find the inverse of that.

$\displaystyle y=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^x}{e^x} \cdot \frac{e^x-e^{-x}}{e^x+e^{-x}}$

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

Let $\displaystyle X=e^{2x}$ (easier to type )

$\displaystyle y=\frac{X-1}{X+1}$

--> $\displaystyle y(X+1)=X-1$

----> $\displaystyle X(y-1)=-1-y$

-------> $\displaystyle \boxed{X=\frac{1+y}{1-y}}$

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

---> $\displaystyle 2x=\ln(1+y)-\ln(1-y)$

------>$\displaystyle \boxed{\color{red} x=\tanh^{-1}(y)=\frac 12 \left(\ln(1+y)-\ln(1-y)\right)}$

Then derivate this...

4. Sorry, my mind is currently completely blank.

$\displaystyle \frac{\mathrm{d} (\mathrm{ln}x)}{\mathrm{d}x} = \frac{1}{x}$

Hence doesn't the derivative of $\displaystyle \frac{1}{2}\ln \frac{1+x}{1-x} = 1 / (\frac{1+x}{1-x}) = \frac{1-x}{1+x}$

5. Originally Posted by Air
Sorry, my mind is currently completely blank.

$\displaystyle \frac{\mathrm{d} (\mathrm{ln}x)}{\mathrm{d}x} = \frac{1}{x}$

Hence doesn't the derivative of $\displaystyle \frac{1}{2}\ln \frac{1+x}{1-x} = 1 / (\frac{1+x}{1-x}) = \frac{1-x}{1+x}$
Don't forget chain rule

$\displaystyle \left(\frac{1}{2}\ln \frac{1+x}{1-x}\right)' = \frac1{\left(\frac{1+x}{1-x}\right)}\frac{d}{dx}\left(\frac{1+x}{1-x}\right)$

6. Transform : $\displaystyle \ln \frac{1+x}{1-x}=\ln(1+x)-\ln(1-x)$

I guess it'd be easier this way

7. Got so close but have made an error in my algebra (I think). Can you spot where I have gone wrong.

$\displaystyle \frac{1}{2}\ln \frac{1+x}{1-x} = \frac{1}{2} \ln (1+x) - \frac{1}{2} \ln (1-x)$

Derivative:
$\displaystyle \frac{1}{2}.\frac{1}{1+x} - \frac{1}{2}.\frac{1}{1-x}$

$\displaystyle \frac{1}{2+2x} - \frac{1}{2-2x}$

$\displaystyle \frac{(2-2x) - (2+2x)}{(2+2x)(2-2x)}$

$\displaystyle \frac{-4x}{-4x^2 +4}$

$\displaystyle \frac{-4x}{4(1-x^2)}$

$\displaystyle \frac{-x}{1-x^2}$

8. Originally Posted by Air

$\displaystyle \frac{1}{2}\ln \frac{1+x}{1-x} = \frac{1}{2} \ln (1+x) - \frac{1}{2} \ln (1-x)$

Derivative:
$\displaystyle \frac{1}{2}.\frac{1}{1+x} - \frac{1}{2}.\frac{1}{1-x}$
You missed the chain rule by takin' the derivative of $\displaystyle \ln(1-x).$

(Typo fixed.)

9. The derivative of $\displaystyle \ln(1-x)$ is $\displaystyle {\color{red}-} \frac{1}{1-x}$, since the derivative of $\displaystyle 1-x$ is -1

10. Originally Posted by Krizalid
You missed the chain rule by takin' the derivative of $\displaystyle \frac1{1-x}.$
Of $\displaystyle \ln(1-x)$, right ?

11. Hi
How about using $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left(f^{-1}(x)\right)=\frac{1}{f'\circ f^{-1}(x)}$ with $\displaystyle f(x)=\tanh x,\,f^{-1}(x)=\tanh^{-1}x,\,f'(x)=\tanh' x=1-\tanh^2x$ ?

If you don't know this, it comes from the derivative of $\displaystyle f\circ f^{-1}(x)=x$.