# Single Variable Calculus, Chapter 7, 7.3-2, Section 7.3-2, Problem 44

Differentiate $\displaystyle y = \frac{e^u - e^{-u}}{e^u + e^{-u}}$

\begin{aligned} y' =& \frac{\displaystyle e^u - \frac{1}{e^u}}{\displaystyle e^u + \frac{1}{e^u}} \\ \\ y' =& \frac{\displaystyle \frac{e^{2u} - 1}{\cancel{e^u}}}{\displaystyle \frac{e^{2u} + 1}{\cancel{e^u}}} \\ \\ y' =& \frac{e^{2u} - 1}{e^{2u} + 1} \\ \\ y' =& \frac{d}{du} \left( \frac{e^{2u} - 1}{e^{2u} + 1} \right) \\ \\ y' =& \frac{\displaystyle (e^{2u} + 1) \frac{d}{du} (e^{2u} - 1) (e^{2u} - 1) \frac{d}{du} (e^{2u} + 1) }{(e^{2u} + 1)^2} \\ \\ y' =& \frac{\displaystyle (e^{2u} + 1) (e^{2u}) \frac{d}{du} (2u) - (e^{2u} - 1) (e^{2u}) \frac{d}{du} (2u) }{(e^{2u} + 1)^2 } \\ \\ y' =& \frac{(e^{2u} + 1) (2e^{2u}) - (e^{2u} - 1) (2e^{2u}) }{(e^{2u} + 1)^2 } \\ \\ y' =& \frac{2e^{2u} [(e^{2u} + 1) - (e^{2u} - 1)] }{(e^{2u} + 1)^2} \\ \\ y' =& \frac{2e^{2u} (\cancel{e^{2u}} + 1 - \cancel{e^{2u}} + 1) }{(e^{2u} + 1)^2} \\ \\ y' =& \frac{2e^{2u} (2)}{(e^{2u + 1})^2} \\ \\ y' =& \frac{4e^{2u}}{(e^{2u} + 1)^2} \end{aligned}

Posted on