Remember that $\cosh(x)=\frac{1}{2}(e^x+e^{-x})$, $\sinh(x)=\frac{1}{2}(e^x-e^{-x})$ and therefore $(\cosh (x))'=\sinh(x)$, $(\sinh(x))'=\cosh(x)$ like fore ordinary $\cos(x)$ and $\sin(x)$ albeit without any minus sign, Further $\cosh^2(x)-\sinh^2(x)=1$.
Then $(\tanh(x))'=\cosh^{-2}(x)$, $(\coth(x))'=\cosh^{-2}(x)$ and so on.
Inverse hyperbolic functions could be expressed through logarithms
http://en.wikipedia.org/wiki/Inverse_hyperbolic_function but to calculate their derivatives one needs just to remember derivatives of "direct" hyperbolic functions and how to differentiate inverse function. Then $(\operatorname{arcosh}(x))'=(x^2-1)^{-1/2}$, $(\operatorname{arsinh}(x))'=(x^2+1)^{-1/2}$, $(\operatorname{artanh}(x))'=(1-x^2)^{-1}$ ($|x|<1$), $(\operatorname{arcoth}(x))'=(1-x^2)^{-1}$ ($|x|>1$).
Even this is more than enough.
