Answer
$-\csc x$
Work Step by Step
We have: $z(x)= \ln |\csc x+\cot x|$
We differentiate both sides with respect to $x$.
Use $\dfrac{d}{dx} [\ln |u|]=\dfrac{1}{u}\dfrac{du}{dx}$
$z^{\prime}(x)=\dfrac{d}{dx} [\ln |\csc x+\cot x|] \\= \dfrac{1}{\csc x+\cot x} \dfrac{d}{dx} [\csc x+\cot x]$
Simplify to obtain:
$z^{\prime}(x)= \dfrac{1}{\csc x+\cot x} (-\csc x \cot x -\csc^2 x)=-\csc x$
