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.

For your convenience, you could first write
$$\frac{1+x}{1-x}=\frac2{1-x}-1\implies\left(\frac{1+x}{1-x}\right)'=\frac2{(1-x)^2}$$
Thus, with the aid of the chain rule:
$$\left(\arctan\sqrt\frac{1+x}{1-x}\right)'=\frac1{1+\frac{1+x}{1-x}}\cdot\frac1{2\sqrt\frac{1+x}{1-x}}\cdot\frac2{(1-x)^2}=$$
$$=\frac{1-x}2\cdot\frac{\sqrt{1-x}}{2\sqrt{1+x}}\cdot\frac2{(1-x)^2}=\frac1{2\sqrt{1-x^2}}$$
