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:
$$g(x) = \frac{1+x}{1-x}$$
$$u = 1+x$$
$$u' = 1$$
$$v = 1-x$$
$$v' = -1$$
$$g'(x) = \frac{(1-x) -(-1)(1+x)}{(1-x)^2}$$
$$g'(x) = \frac{1-x+1+x}{(1-x)^2}$$
$$g'(x) = \frac{2}{(1-x)^2}$$
$$y' = \frac{1}{2}(\frac{1+x}{1-x})^{-\frac{1}{2}}(\frac{2}{(1-x)^2})
$$

$$
y' = \frac{d}{dx}\left(\sqrt{\frac{1+x}{1-x}}\right)
$$
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
$$
y' = \sqrt{\frac{1-x}{1+x}}\frac{1}{(1-x)^2}
$$
