Answer
$y'=\csc x$
Work Step by Step
$y=\ln (\csc x-\cot x)\\
y'=\frac{1}{\csc x-\cot x} \times (-\cot x\csc x)-(-(\csc x)^2) \\
y'=\frac{1}{\csc x-\cot x} \times \csc ^{2} x-\cot x \csc x \\
y'=\frac{\csc x(\csc x-\cot x)}{\csc x-\cot x} \\
y'=\csc x$
