<s>[INST] Derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ I am trying to find the derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ by only using the formula $\arctan\left(u\left(x\right)\right)'=\frac{u'\left(x\right)}{u\left(x\right)^2+1}$. I don't honestly understand this formula, here are however my calculations. Is this the right way to find the derivative? And how do I proceed?
$$\LARGE
\frac{\frac{1}{2\sqrt{\frac{1+x}{1-x}}}\cdot \frac{\left(1-x\right)-\left(1+x\right)}{x^2-2x+1}}{\frac{x+1}{1-x}+1}
$$
I also appreciate if you can explain what that "formula" exactly is.
 [/INST] Well done! Your answer is correct. To make it a bit more presentable, let us simplify this further to get: $$f’(x) = \frac{\frac{\sqrt{\frac{1-x}{1+x}}}{2}\times \frac{-2x}{x^2-2x+1}}{2}(1-x)$$ $$=-\frac{(1-x)^{\frac32}}{2\sqrt{1+x}(x^2-2x+1)}$$ $$=-\frac{1}{2\sqrt{1-x^2}}$$

EDIT:
Well, you have your derivative calculated wrong, it should be $$\frac{(1-x)-(1+x)(-1)}{x^2-2x+1} = \frac{2}{x^2-2x+1}$$
So, removing the minus sign, we have: $$f’(x) = \frac{1}{2\sqrt{1-x^2}}$$
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