Answer
$$\frac{d}{d x} \operatorname{sech} x=-\operatorname{sech} x \tanh x$$
Work Step by Step
Since
$$ \operatorname{sech} x =\frac{2}{e^x+e^{-x}}$$
Then
\begin{align*}
\frac{d}{d x} \operatorname{sech} x&= \frac{d}{d x}\frac{2}{e^x+e^{-x}}\\
&= 2\frac{d}{dx}\left(\left(e^x+e^{-x}\right)^{-1}\right)\\
&=-\frac{2\left(e^{x}-e^{-x}\right)}{\left(e^{x}+e^{-x}\right)^{2}}\\
&= -\frac{ \left(e^{x}-e^{-x}\right)}{\left(e^{x}+e^{-x}\right) } \frac{2}{\left(e^{x}+e^{-x}\right) }\\
&= -\operatorname{sech} x \tanh x
\end{align*}
