The quotient rule states that if $y=\frac{u}{v}$, then
$$y'=\frac{u'v-uv'}{v^2}$$
where $u'$ and $v'$ are the derivatives of $u$ and $v$, respectively.
In our case, we have $u=\operatorname{sech} x$ and $v=1+\cosh x$. Therefore,
$$u'=-\operatorname{sech} x
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