Answer
$$\frac{{dy}}{{dx}} = \frac{{2 + 5\cosh \left( {5x} \right)\sinh \left( {5x} \right)}}{{\sqrt {4x + {{\cosh }^2}\left( {5x} \right)} }}$$
Work Step by Step
$$\eqalign{
& y = \sqrt {4x + {{\cosh }^2}\left( {5x} \right)} \cr
& {\text{find the derivatvive}} \cr
& y' = \left( {\sqrt {4x + {{\cosh }^2}\left( {5x} \right)} } \right)' \cr
& {\text{chain rule}} \cr
& y' = \frac{1}{2}{\left( {4x + {{\cosh }^2}\left( {5x} \right)} \right)^{ - 1/2}}\left( {4x + {{\cosh }^2}\left( {5x} \right)} \right)' \cr
& {\text{use the theorem 6}}{\text{.8}}{\text{.3 }}\left( {{\text{ see page 476}}} \right){\text{ }} \cr
& \frac{d}{{dx}}\left[ {\cosh u} \right] = \sinh u\frac{{du}}{{dx}} \cr
& y' = \frac{1}{2}{\left( {4x + {{\cosh }^2}\left( {5x} \right)} \right)^{ - 1/2}}\left( {4 + 2\cosh \left( {5x} \right)\sinh \left( {5x} \right)\left( {5x} \right)'} \right) \cr
& y' = \frac{1}{2}{\left( {4x + {{\cosh }^2}\left( {5x} \right)} \right)^{ - 1/2}}\left( {4 + 10\cosh \left( {5x} \right)\sinh \left( {5x} \right)} \right) \cr
& {\text{simplifying}} \cr
& y' = {\left( {4x + {{\cosh }^2}\left( {5x} \right)} \right)^{ - 1/2}}\left( {2 + 10\cosh \left( {5x} \right)\sinh \left( {5x} \right)} \right) \cr
& y' = \frac{{4 + 10\cosh \left( {5x} \right)\sinh \left( {5x} \right)}}{{2\sqrt {4x + {{\cosh }^2}\left( {5x} \right)} }} \cr
& \frac{{dy}}{{dx}} = \frac{{2 + 5\cosh \left( {5x} \right)\sinh \left( {5x} \right)}}{{\sqrt {4x + {{\cosh }^2}\left( {5x} \right)} }} \cr} $$
