## Calculus (3rd Edition)

$$y'= -sech\, x-sech\, x \, csch^2x.$$
Since $y=sech x\coth x$, then the derivative, by using the product rule $(uv)'=uv'+u'v$ and the facts that $(\ sech \ x)= -\ sech \ x \tanh x$ and $(\coth x)'=-csch^2 x$, is given by $$y'= -sech\, x\tanh x\coth x-sech\, x \, csch^2x\\ y'= -sech\, x-sech\, x \, csch^2x$$