Answer
$$\frac{{dy}}{{dx}} = \frac{{{e^{2x}} - {e^{ - 2x}}}}{{\sqrt {{e^{2x}} + {e^{ - 2x}}} }}$$
Work Step by Step
$$\eqalign{
& y = \sqrt {{e^{2x}} + {e^{ - 2x}}} \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\sqrt {{e^{2x}} + {e^{ - 2x}}} } \right] \cr
& {\text{Use }}\frac{d}{{dx}}\left[ {\sqrt u } \right] = \frac{1}{{2\sqrt u }}\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {{e^{2x}} + {e^{ - 2x}}} }}\frac{d}{{dx}}\left[ {{e^{2x}} + {e^{ - 2x}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {{e^{2x}} + {e^{ - 2x}}} }}\left( {2{e^{2x}} - 2{e^{ - 2x}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{{e^{2x}} - {e^{ - 2x}}}}{{\sqrt {{e^{2x}} + {e^{ - 2x}}} }} \cr} $$
