Answer
$$ - \operatorname{sech} x\tanh x$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left( {\operatorname{sech} x} \right) = - \operatorname{sech} x\tanh x \cr
& {\text{ hyperbolic function for secant}} \cr
& \frac{d}{{dx}}\left( {\operatorname{sech} x} \right) = \frac{d}{{dx}}\left( {\frac{1}{{\cosh x}}} \right) \cr
& {\text{differentiate by the product rule}} \cr
& \frac{{\cosh x\frac{d}{{dx}}\left[ 1 \right] - 1\frac{d}{{dx}}\left[ {\cosh x} \right]}}{{{{\left( {\cosh x} \right)}^2}}} \cr
& \frac{{\cosh x\left( 0 \right) - 1\left( {\sinh x} \right)}}{{{{\left( {\cosh x} \right)}^2}}} \cr
& {\text{simplify}} \cr
& \frac{{ - \sinh x}}{{{{\left( {\cosh x} \right)}^2}}} \cr
& or \cr
& - \frac{1}{{\cosh x}}\frac{{\sinh x}}{{\cosh x}} \cr
& {\text{hyperbolic functions}} \cr
& - \operatorname{sech} x\tanh x \cr} $$
