<s>[INST] 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:
$$g(x) = \frac{1+x}{1-x}$$
$$u = 1+x$$
$$u' = 1$$
$$v = 1-x$$
$$v' = -1$$
$$g'(x) = \frac{(1-x) -(-1)(1+x)}{(1-x)^2}$$
$$g'(x) = \frac{1-x+1+x}{(1-x)^2}$$
$$g'(x) = \frac{2}{(1-x)^2}$$
$$y' = \frac{1}{2}(\frac{1+x}{1-x})^{-\frac{1}{2}}(\frac{2}{(1-x)^2})
$$
 [/INST] 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 
$$ y' = \sqrt{ \frac{1-x}{1+x} } (1-x)^2$$
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