Differentiate the function: $y=\frac{e^u-e^{-u}}{e^u+e^{-u}}$ $$y=\frac{e^u-e^{-u}}{e^u+e^{-u}}$$
Using the Quotient Rule, 
$$\frac{e^u+e^{-u}[e^u-e^{-u}]'-[(e^u-e^{-u})[e^u+e^{-u}]']}{(e^u+e^{-u})^2}$$
 A: There are only two letters in the right-hand side of your equation. $e$ is a constant, and $u$ is a variable. $e$ is a specific constant, the base of the natural logarithms and sometimes called Euler's number, so I can't think of a general rule that applies to it. $x,y,z,u,v$ are usually variables in calculus.
$$\begin{align}
\left(\frac{e^u-e^{-u}}{e^u+e^{-u}}\right)'
 &=\frac{(e^u-e^{-u})'(e^u+e^{-u})-(e^u+e^{-u})'(e^u-e^{-u})}{(e^u+e^{-u})^2} \\[2ex]
 &=\frac{(e^u+e^{-u})(e^u+e^{-u})-(e^u-e^{-u})(e^u-e^{-u})}{(e^u+e^{-u})^2} \\[2ex]
 &=\frac{[(e^u)^2+2e^ue^{-u}+(e^{-u})^2]-[(e^u)^2-2e^ue^{-u}+(e^{-u})^2]}{} \\[2ex]
 &=\frac{[e^{2u}+2+e^{-2u}]-[e^{2u}-2+e^{-2u}]}{(e^u+e^{-u})^2} \\[2ex]
 &= \frac 4{(e^u+e^{-u})^2}
\end{align}$$
By the way, if you know the hyperbolic functions this shows that
$$(\tanh u)'=\operatorname{sech}^2 u$$
A: Another alternative would be logarithmic differentiation. $$\ln y = \ln (e^u - e^{-u}) - \ln (e^u + e^{-u}).$$ Differentiating implicitly with respect to $u$ yields $$\frac{1}{y} \frac{\mathrm{d}y}{\mathrm{d}u} = \frac{e^u + e^{-u}}{e^u - e^{-u}} - \frac{e^u - e^{-u}}{e^u + e^{-u}}$$
Now multiplying both sides by $y$ yields $$\begin{align}\frac{\mathrm{d}y}{\mathrm{d}u} &= \frac{e^u-e^{-u}}{e^u+e^{-u}} \left(\frac{e^u + e^{-u}}{e^u - e^{-u}} - \frac{e^u - e^{-u}}{e^u + e^{-u}}\right) \\ \\ &= 1 - \left(\frac{e^u - e^{-u}}{e^u + e^{-u}}\right)^2\end{align}$$
A: From Euler's formula, the function you wrote is simply $\tanh u$, and just like $\tan'u=1+\tan^2u$ $=\dfrac1{\cos^2u}=\sec^2u,~$ in a similar manner, $~\tanh'u=1-\tanh^2u=\dfrac1{\cosh^2u}=\text{sech}^2u$.
