Answer
$$ y'=-\frac{csch^2c}{\coth x} .$$
Work Step by Step
Recall that $(\ln x)'=\dfrac{1}{x}$
Recall that $(\coth x)'=-\csc^2 x$
Since $ y=\ln(\coth x)$, then the derivative, by using the chain rule, is given by
$$ y'=\frac{1}{\coth x} (\coth x)'=-\frac{csch^2c}{\coth x} .$$
