<s>[INST] First derivative meaning in this case If we have a function:
$$f(x)=\frac{x}{2}+\arcsin{\frac{2x}{1+x^2}}$$
And it's first derivative is calculated as:
$$f'(x)=\frac{1}{2}+\frac{1}{\sqrt{1-\big(\frac{2x}{1+x^2}\big)^2}}\frac{2+2x^2-4x^2}{(1+x^2)^2}=$$
$$\frac{1}{2}+\frac{2(1-x^2)}{\sqrt{\frac{(1-x^2)^2}{(1+x^2)^2}}\cdot(1+x^2)^2}=$$
$$\frac{1}{2}+\frac{2(1-x^2)}{|1-x^2|\cdot(1+x^2)}=$$
$$\frac{1}{2}+\frac{2\cdot sgn(1-x^2)}{1+x^2}$$
Why did my teacher say the critical points are at $x=\pm1$? Then she wrote:
$$f_+'(1)=-\frac{1}{2}$$
$$f_-'(1)=\frac{3}{2}$$
and I'm not sure what she meant by that eather, nor by the following:
$$f'(x)=
\begin{cases}
\frac{x^2-3}{2(x^2+1)}, & |x|>1 \\
\frac{x^2+5}{2(x^2+1)}, & |x|<1
\end{cases}$$
Thank you for your time.
 [/INST] If $\arctan x=y,x=\tan y$ and $-\dfrac\pi2\le y\le\dfrac\pi2$
(See definition of the principal values of inverse trigonometric functions)
$$\dfrac{2x}{1+x^2}=\cdots=\sin2y$$
$$\implies\arcsin\dfrac{2x}{1+x^2}=\arcsin(\sin2y)=\begin{cases}2y&\mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2\iff-1\le x\le1\\
\pi-2y & \mbox{if }2y>\dfrac\pi2\\
-\pi-2y  & \mbox{if }2y<-\dfrac\pi2 \end{cases} $$
From this relation, we can easily discern the change of sign of $$\dfrac{d\left(\arcsin\dfrac{2x}{1+x^2}\right)}{dx}$$ for the different ranges of values of $x$
</s>