Answer
$$ - {\operatorname{csch} ^2}x$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left( {\cot x} \right) = - {\operatorname{csch} ^2}x \cr
& {\text{ hyperbolic function for cotangent}} \cr
& \frac{d}{{dx}}\left( {\frac{{\cosh x}}{{\sinh x}}} \right) \cr
& {\text{differentiate by the product rule}} \cr
& \frac{{\sinh x\frac{d}{{dx}}\left[ {\cosh x} \right] - \cosh x\frac{d}{{dx}}\left[ {\sinh x} \right]}}{{{{\left( {\sinh x} \right)}^2}}} \cr
& \frac{{\sinh x\left( {\sinh x} \right) - \cosh x\left( {\cosh x} \right)}}{{{{\sinh }^2}x}} \cr
& {\text{simplify}} \cr
& \frac{{{{\sinh }^2}x - {{\cosh }^2}x}}{{{{\sinh }^2}x}} \cr
& or \cr
& - \frac{{{{\cosh }^2}x - {{\sinh }^2}x}}{{{{\sinh }^2}x}} \cr
& {\text{hyperbolic identity}} \cr
& - \frac{1}{{{{\sinh }^2}x}} \cr
& - {\operatorname{csch} ^2}x \cr} $$
