Differentiate an inverse tangent and a logarithmic square root

The chain rule gives $\dfrac{-2/x^2}{1+(2/x)^2}=-\dfrac2{x^2+4}$, on the original domain $x\ne0$.

On the real domain $-1<x<1$, rewrite the expression as $\frac12[\ln(1+x)-\ln(1-x)]$. Its derivative is $\frac12[1/(1+x)+1/(1-x)]=\dfrac1{1-x^2}$.

The source labels this longer route not recommended. Let $u=(1+x)/(1-x)$, so $u\prime=2/(1-x)^2$. Then $\dfrac{d}{dx}\ln\sqrt u=\dfrac1{\sqrt u}\dfrac{u\prime}{2\sqrt u}=\dfrac{u\prime}{2u}=\dfrac1{(1-x)(1+x)}$.