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

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


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\begin{equation}
\begin{aligned}

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





\end{aligned}
\end{equation}
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