Answer
$f'\left( x \right) = {\operatorname{sech} ^3}x - \operatorname{sech} x{\tanh ^2}x$
Work Step by Step
$$\eqalign{
& y = \operatorname{sech} x\tanh x \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\operatorname{sech} x\tanh x} \right] \cr
& {\text{Use the product rule}} \cr
& f'\left( x \right) = \operatorname{sech} x\frac{d}{{dx}}\left[ {\tanh x} \right] + \tanh x\frac{d}{{dx}}\left[ {\operatorname{sech} x} \right] \cr
& {\text{Use the Derivatives of Hyperbolic Functions }} \cr
& f'\left( x \right) = \operatorname{sech} x\left( {{{\operatorname{sech} }^2}x} \right) + \tanh x\left( { - \operatorname{sech} x\tanh x} \right) \cr
& {\text{Simplify}} \cr
& f'\left( x \right) = {\operatorname{sech} ^3}x - \operatorname{sech} x{\tanh ^2}x \cr} $$
