absolute value in derivative of inverse hyperbolic cosecant derivative of inverse hyperbolic cosecant is: $$\frac {-1} {|x|\sqrt{1+x^2}}$$ i saw in some website the absolute value of $x$ (in denominator) obtained after considering both $x>0$ and $x<0$. but, i don't have idea how. here i'll attach result from both cases when $x>0$ the derivative : $$\frac {-1} {x\sqrt{1+x^2}}$$ when $x<0$ the derivative : $$\frac {1} {x\sqrt{1+x^2}}$$ first,how both can turn into the general function (most top function that give x an absolute value) ? second, i saw derivative of inverse hyperbolic secant, and it done by similiar ways, but absolute value is nowhere to be found. how ? thankyou in advanced

The first one is because $$\text{csch}^{-1}(-x)=-\text{csch}^{-1}(x)$$ On the other side, in the real domain, $\text{sech}^{-1}(x)$ is undefined.