The derivation in the previous section carries over virtually unchanged to hyperbolic trigonometric functions. Recall from Section 4.2 the geometric definition of the hyperbolic trigonometric functions in (Equations (2) and (3) of §4.2) as shown in Figure 4.2.

We compute the derivatives of these functions using the same technique as before. We choose to work on the hyperbola $y^2-x^2=\rho^2$, that is, the hyperbola shown in Figure 4.2 that contains the point $B$. What do we know? We know that (infinitesimal) arclength along the hyperbola is given by \begin{equation} ds = \rho\,d\beta \end{equation} but we also have the (infinitesimal, Lorentzian) Pythagorean Theorem, which tells us that \begin{equation} ds^2 = dx^2 - dy^2 \end{equation} Furthermore, from $y^2-x^2=\rho^2$, we obtain \begin{equation} x\,dx = y\,dy \label{hdiff} \end{equation} Putting this information together, we have \begin{equation} \rho^2\,d\beta^2 = dx^2 - dy^2 = dx^2 \left( 1 - \frac{x^2}{y^2} \right) = \rho^2\,\frac{dx^2}{y^2} \end{equation} so that \begin{equation} d\beta^2 = \frac{dx^2}{y^2} = \frac{dy^2}{x^2} \end{equation} where we have used (\ref{hdiff}) in the last step. Carefully using Figure \ref{hyperfig1} to check signs, we can take the square root and rearrange terms to obtain \begin{eqnarray} dy &=& x\,d\beta \nonumber\\ dx &=& y\,d\beta \end{eqnarray} Finally, (Equations (2) and (3) of §4.2) and using the fact that $\rho=\hbox{constant}$, we obtain \begin{eqnarray} d\sinh\beta &=& \cosh\beta\,d\beta \nonumber\\ d\cosh\beta &=& \sinh\beta\,d\beta \end{eqnarray} We have thus determined the derivatives of the basic hyperbolic trigonometric functions.