<s>[INST] Let $f(x)=\arctan x -\frac{ln|x|}{2}$. Then $f(x)$ is increasing in? Let $x<0$. $f'(x)=\frac{1}{1+x^2}+\frac{1}{2x}>0$
$$f'(x)=\frac{(1+x)^2}{2x(1+x^2)}$$ which is always negative. But answer is $x\in (-\infty,0)$.
 [/INST] A:  The derivative of $\ln |x|=\frac 1 x$ for any $x\ne 0$
Proof:  for x<0,  $\ln |x|$=$\ln (-x)$  derivative by the chain rule is $\frac {du} {u}$, so $f'(x)=\frac {-1} {-x}=\frac 1 x$
So,  $f'(x)=\frac {1} {1+x^2} - \frac {1} {2x}$,  which when $x<0$, we have both terms are positive, so $f'(x)>0$ when $x<0$.  However, when $x>0$,  we need to consider which is bigger, $\frac {1} {1+x^2}$  or $\frac {1} {2x}$.  The larger fraction is going to be the one with the smaller denominator, so we can compare denominators to make our life easier.  consider $g(x)=1+x^2-2x=x^2-2x+1=(x-1)^2$.
This is always nonnegative, which means $1+x^2\ge 2x$. for $x\ge 0$], so $\frac {1} {1+x^2}\le \frac 1 {2x}$,  hence $f'(x)\le 0$ when $x>0$,  thus we only have increasing when $x\in (-\infty, 0)$
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