Answer
$$\frac{{dy}}{{dx}} = - 4\operatorname{sech} \left( {4x - 1} \right)\tanh \left( {4x - 1} \right)$$
Work Step by Step
$$\eqalign{
& y = \operatorname{sech} \left( {4x - 1} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\operatorname{sech} \left( {4x - 1} \right)} \right] \cr
& {\text{Using theorem 5}}{\text{.8}} \cr
& \frac{{dy}}{{dx}} = - \left[ {\operatorname{sech} \left( {4x - 1} \right)\tanh \left( {4x - 1} \right)} \right]\frac{d}{{dx}}\left[ {\left( {4x - 1} \right)} \right] \cr
& \frac{{dy}}{{dx}} = - \operatorname{sech} \left( {4x - 1} \right)\tanh \left( {4x - 1} \right)\left( 4 \right) \cr
& \frac{{dy}}{{dx}} = - 4\operatorname{sech} \left( {4x - 1} \right)\tanh \left( {4x - 1} \right) \cr} $$
