Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 352: 50

$y'=\frac{2e^{2x}}{(e^{2x}+1)^2}$

$y=\frac{e^{2x}}{e^{2x}+1}$ $u=e^{2x}$ $u’=2e^{2x}$ $v=e^{2x}+1$ $v’=2e^{2x}$ $y’=\frac{(2e^{2x})(e^{2x}+1)-e^{2x})(2e^{2x})}{(e^{2x}+1)^2}$ $=\frac{(2e^{2x})(e^{2x}+1-e^{2x})}{(e^{2x}+1)^2}$ $=\frac{2e^{2x}}{(e^{2x}+1)^2}$