How to differentiate $y=\sqrt{\frac{1+x}{1-x}}$? I'm trying to solve this problem but I think I'm missing something. 
Here's what I've done so far:
$$g(x) = \frac{1+x}{1-x}$$
$$u = 1+x$$
$$u' = 1$$
$$v = 1-x$$
$$v' = -1$$
$$g'(x) = \frac{(1-x) -(-1)(1+x)}{(1-x)^2}$$
$$g'(x) = \frac{1-x+1+x}{(1-x)^2}$$
$$g'(x) = \frac{2}{(1-x)^2}$$
$$y' = \frac{1}{2}(\frac{1+x}{1-x})^{-\frac{1}{2}}(\frac{2}{(1-x)^2})
$$
 A: Make your life simpler by logging: 
$$
L f(x) = \frac{1}{2} \log (1+x) - \frac{1}{2} \log (1-x)
$$
Now differentiate this and multply by $f(x)$. 
A: Another idea:
$$\frac{1+x}{1-x}=\frac2{1-x}-1\implies \left(\frac{1+x}{1-x}\right)'=\frac2{(1-x)^2}\implies$$
$$\left(\sqrt\frac{1+x}{1-x}\right)'=\sqrt\frac{1-x}{1+x}\cdot\frac1{(1-x)^2}$$
Added on request:
$$\sqrt\frac{1-x}{1+x}\cdot\frac1{(1-x)^2}=\frac1{\sqrt{1+x}}\cdot\frac1{\sqrt{1-x}\cdot(1-x)}=\frac1{\sqrt{1-x^2}(1-x)}$$
A: let 
$x=\cos u$
$$y=\sqrt{\frac{1+x}{1-x}}=\sqrt{\frac{1+\cos u}{1-\cos u}}=\cot(\frac{u}{2})$$
$$y'=\frac{-1}{2}\csc^2(u/2)$$
then 
$u=\arccos x$
A: Write: 
$$ y^2 = \frac{ 1 + x}{1 -x } \iff y^2 = 1 + \frac{2x}{1-x}$$ 
so taking derivative with repect to $x$ gives 
$$ 2y y' = \frac{2(1-x) + 2x}{(1-x)^2} =\frac{2}{(1-x)^2} $$
so 
$$ y' = \sqrt{ \frac{1-x}{1+x} } (1-x)^2$$
A: $$ y^2=\frac{1+x}{1-x}=\frac2{1-x}-1\implies Derivative =\frac2{(1-x)^2} $$
$$ 2 y y ^{'} = RHS $$
$$y'=\sqrt\frac{1-x}{1+x}\cdot\frac1{(1-x)^2}$$
A: $$
y' = \frac{d}{dx}\left(\sqrt{\frac{1+x}{1-x}}\right)
$$
Chain rule through the square root:
$$
y' = \frac{1}{2}\sqrt{\frac{1-x}{1+x}}\frac{d}{dx}\left(\frac{1+x}{1-x}\right)
$$
Quotient rule on what's left
$$
y' = \frac{1}{2}\sqrt{\frac{1-x}{1+x}}\left(\frac{1-x-(1+x)(-1)}{(1-x)^2}\right)
$$
Simplify
$$
y' = \sqrt{\frac{1-x}{1+x}}\frac{1}{(1-x)^2}
$$
