Problem derivating the inverse cossecant $\DeclareMathOperator{\arccsc}{arccsc}$
I've tried to derivate the $\arccsc$ function but something seems to be wrong with my reasoning.
$$\csc (\arccsc x) = x $$
$$\frac{d}{dx} \csc (\arccsc x) = \frac{d}{dx} x = 1 $$
As $\frac{d}{dx} \csc x = - \cot x \csc x$ by the Chain Rule, one gets:
$$ -\cot(\arccsc(x)) \csc(\arccsc(x)) \frac{d}{dx} \arccsc(x) = 1 $$
$$ \frac{d}{dx} \arccsc(x) = -\frac{1}{x \cot(\arccsc(x))} $$
But $\cot^2 x + 1 = \csc^2 x $, so:
$$ \frac{d}{dx} \arccsc(x) = -\frac{1}{x \sqrt{\csc^2(\arccsc(x)) - 1}} = -\frac{1}{x \sqrt{x^2 -1 }}  $$
But it should be:
$$ \frac{d}{dx} \arccsc(x) = -\frac{1}{|x| \sqrt{x^2 -1 }} $$
What step is wrong?
 A: $\DeclareMathOperator{\arccsc}{arccsc}$You have to be careful of the branch you are picking for these functions, and exactly which $x$ are actually in the domain.  Your derivation is correct for one particular choice of branch for the $\arccsc$, but not the one your text is using.
As for where, more specifically, your argument hits a snag, it's in your application of $\cot^2(x)=\csc^2(x)-1$.  You took the square root, but failed to take care with the signs.  In general taking the square root yields $|\cot(x)|=\sqrt{\csc^2(x)-1}$.  
Here's where the choice of branch for $\arccsc$ comes in.  The choice of branch of the $\arccsc(x)$ in your case puts the angle in the interval $(-\pi/2,\pi/2)$.  On this interval the $\cot$ function changes sign.  It's positive on $(-\pi/2,0)$, but negative on $(0,\pi/2)$.  Thus you pick up an additional $\operatorname{sgn}(x)$ in the denominator, which gives you $|x|$ instead of $x$.
You can also see graphically why your answer is flawed by looking at the graph for the $\arccsc$, with codomain $(-\pi/2,\pi/2)$, and noticing that it is strictly decreasing.  Your originally derived formula is increasing for $x<-1$.
