Answer
$$\frac{{dy}}{{dx}} = \tanh x$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {\cosh x} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln \left( {\cosh x} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\cosh x}}\frac{d}{{dx}}\left[ {\cosh x} \right] \cr
& {\text{Using the rules of theorem 5}}{\text{.18 }}\left( {{\text{Page 385}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\cosh x}}\left( {\sinh x} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{\sinh x}}{{\cosh x}} \cr
& {\text{Simplify}} \cr
& \frac{{dy}}{{dx}} = \tanh x \cr} $$
