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.
 A: 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: $$f’(x) = \frac{1}{2\sqrt{1-x^2}}$$
A: The chain rule says that, if $u$ is some differentiable function of $x$, the composition $\,\arctan u(x)$ has derivative:
\begin{align}
\bigl(\arctan u(x)\bigr)'_x&=\bigl(\arctan u(x)\bigr)'_u\cdot u'_x(x)=\frac1{1+u^2(x)}\cdot u'_x(x)=\frac1{1+\smash[b]{\dfrac{1+x}{1-x}}}\cdot u'_x(x)\\
&=\frac12(1-x)u'_x(x).
\end{align}
Now $u(x)$ is itself a composition: $u(x)=\sqrt{v(x)}$, where $\,v(x)=\dfrac{1+x}{1-x}$, hence
$$u'_x(x)=u(x)'_v\cdot v'_x(x)=\frac1{2\sqrt{v(x)}}\cdot\frac2{(1-x)^2}=\frac{\sqrt{1-x}}{\sqrt{1+x}}\cdot\frac1{(1-x)^2}=\frac1{\sqrt{(1-x)^2(1-x^2})},$$
so that, after simplifying,
$$f'(x)=\frac1{2\sqrt{1-x^2}}.$$
A: 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:
$$\left(\arctan\sqrt\frac{1+x}{1-x}\right)'=\frac1{1+\frac{1+x}{1-x}}\cdot\frac1{2\sqrt\frac{1+x}{1-x}}\cdot\frac2{(1-x)^2}=$$
$$=\frac{1-x}2\cdot\frac{\sqrt{1-x}}{2\sqrt{1+x}}\cdot\frac2{(1-x)^2}=\frac1{2\sqrt{1-x^2}}$$
A: 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*}
A: The formula tells you how to compute the derivative of an $\arctan$ expression.
Let us first focus on $\arctan x=\tan^{-1}x$ and use the inversion formula.
$$y=\arctan x\iff x=\tan y$$ and
$$\frac{dy}{dx}=\frac1{\dfrac{dx}{dy}}=\frac1{\dfrac1{\cos^2y}}=\frac1{\tan^2y+1}=\frac1{x^2+1}.$$
Now by the chain rule,
$$\left(\arctan u(x)\right)'=\frac{d\arctan u}{du}\frac{du}{dx}=\frac{u'(x)}{u^2(x)+1}.$$
