Answer
$\frac{{dy}}{{dx}} = - {\operatorname{sech} ^2}x\operatorname{sech} \left( {\tanh x} \right)\tanh \left( {\tanh x} \right)$
Work Step by Step
$$\eqalign{
& y = \operatorname{sech} \left( {\tanh x} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\operatorname{sech} \left( {\tanh x} \right)} \right] \cr
& {\text{Use the Derivatives of Hyperbolic Functions }} \cr
& \frac{d}{{dx}}\left[ {\operatorname{sech} u} \right] = - \operatorname{sech} u\tanh u\frac{{du}}{{dx}},{\text{ let }}u = \tanh x,{\text{ so}} \cr
& \frac{{dy}}{{dx}} = - \operatorname{sech} \left( {\tanh x} \right)\tanh \left( {\tanh x} \right)\frac{d}{{dx}}\left[ {\tanh x} \right] \cr
& {\text{Differentiate }}\frac{d}{{dx}}\left[ {\tanh x} \right] \cr
& \frac{{dy}}{{dx}} = - \operatorname{sech} \left( {\tanh x} \right)\tanh \left( {\tanh x} \right)\left( {{{\operatorname{sech} }^2}x} \right) \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = - {\operatorname{sech} ^2}x\operatorname{sech} \left( {\tanh x} \right)\tanh \left( {\tanh x} \right) \cr} $$
