Answer
$$\frac{d}{{dx}}\left[ {\operatorname{sech} x} \right] = - \operatorname{sech} x\tanh x$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left[ {\operatorname{sech} x} \right] \cr
& {\text{Using hyperbolic identities}} \cr
& \frac{d}{{dx}}\left[ {\frac{1}{{\cosh x}}} \right] \cr
& {\text{By the definition of }}\cosh x{\text{ }} \cr
& \frac{d}{{dx}}\left[ {\frac{2}{{{e^x} + {e^{ - x}}}}} \right] \cr
& {\text{Differentiating and simplifying}} \cr
& \frac{d}{{dx}}\left[ {2{{\left( {{e^x} + {e^{ - x}}} \right)}^{ - 1}}} \right] \cr
& = - 2{\left( {{e^x} + {e^{ - x}}} \right)^{ - 2}}\frac{d}{{dx}}\left[ {{e^x} + {e^{ - x}}} \right] \cr
& = - 2{\left( {{e^x} + {e^{ - x}}} \right)^{ - 2}}\left( {{e^x} - {e^{ - x}}} \right) \cr
& = - \left( {\frac{2}{{{e^x} + {e^{ - x}}}}} \right)\left( {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} \right) \cr
& {\text{Where }}\tanh x = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}{\text{ and }}\operatorname{sech} x = \frac{2}{{{e^x} + {e^{ - x}}}} \cr
& = - \operatorname{sech} x\tanh x \cr
& {\text{QED}}{\text{.}} \cr} $$
