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.

We have 
\begin{eqnarray*}
\frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2}
\end{eqnarray*}
and
\begin{eqnarray*}
\frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}}
\end{eqnarray*}
and
\begin{eqnarray*}
\frac{d}{dx} \frac{1+x}{1-x} = \frac{ (1-x)--(1+x)}{(1-x)^2}= \frac{2 }{(1-x)^2}.
\end{eqnarray*}
Now we need to do a $3-$ fold chain rule, 
\begin{eqnarray*}
\frac{d}{dx} f(g(h(x))) = h'(x) g'(h(x)) f'(g(h(x))).
\end{eqnarray*}
So in your case we get
\begin{eqnarray*}
\frac{d}{dx} \tan^{-1} \left( \sqrt{\frac{1+x}{1-x}} \right) = \frac{\frac{2 }{(1-x)^2} \frac{1}{2\sqrt{\frac{1+x}{1-x}} }}{1+ \frac{1+x}{1-x}} = \frac{1}{2 \sqrt{1-x^2}}.
\end{eqnarray*}
