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 $\displaystyle \frac{1}{1-x^2}$.
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...
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:
$\displaystyle \frac{1}{2}.\frac{1}{1+x} - \frac{1}{2}.\frac{1}{1-x}$
$\displaystyle \frac{1}{2+2x} - \frac{1}{2-2x}$
$\displaystyle \frac{(2-2x) - (2+2x)}{(2+2x)(2-2x)}$
$\displaystyle \frac{-4x}{-4x^2 +4}$
$\displaystyle \frac{-4x}{4(1-x^2)}$
$\displaystyle \frac{-x}{1-x^2}$
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 $\displaystyle f\circ f^{-1}(x)=x$.