If \(u=\ln (\sec \phi+\tan \phi),\) then \(\phi\) is a function of \(u\) called the Gudermannian of \(u\) \(\phi=\operatorname{gd} u .\) Prove that: $$u=\ln \tan \left(\frac{\pi}{4}+\frac{\phi}{2}\right), \quad \tan \mathrm{gd} u=\sinh u, \quad \sin \mathrm{gd} u=\tanh u, \quad \frac{d}{d u} \operatorname{gd} u=\operatorname{sech} u$$

Trigonometric identities are fundamental tools used to simplify and manipulate expressions involving trigonometric functions.
In the exercise, we encounter identities such as the one linking secant and tangent to exponential forms.
This simplifies our calculations significantly.
For instance, remember that \( sec \phi + tan \phi \) can be re-expressed using the identity: \(( \cos \phi)^{-1}(1 + \sin \phi).\)
This is crucial when working with \( u = \ln(\sec \phi + \tan \phi).\)
Such identities help in transitioning between different forms of representation, making complex expressions more manageable.
To further simplify, we used the half-angle formulas like: \(\tan(\frac{\pi}{4} + \frac{\phi}{2}) = \frac{1 + \sin \phi}{\romanchar{10}\cos \phi}\)

Hyperbolic functions, much like trigonometric functions, are essential in various fields of mathematics and physics.
They are defined using exponential functions and include \( \sinh u, \cosh u, \;\text{and}\; \tanh u.\)
In the exercise, we linked the Gudermannian function to hyperbolic functions.
For example, we proved \( \tan(\text{gd} u) = \sinh u.\)
To do this, recall the definitions of hyperbolic functions: \( \sinh u = \frac{e^u - e^{-u}}{2} \) and \(\cosh u = \frac{e^u + e^{-u}}{2}.\)
Using the identity \( \cosh^2 u - \sinh^2 u = 1,\)
we simplified and related these forms to the Gudermannian function. This interplay between trigonometric and hyperbolic functions demonstrates their intrinsic connections.
These relationships are instrumental in bridging the gap between circular and hyperbolic geometry, leading to more profound insights and applications.

Differentiation is a critical concept in calculus, allowing us to find the rate of change of functions.
In the context of this exercise, we differentiated the Gudermannian function.
Specifically, we showed that \(\frac{d}{du}(\text{gd} u) = \operatorname{sech} u).\)
To achieve this, we utilized the identity \( \sin(\text{gd} u) = \tanh u.\)
Remember that the hyperbolic secant function, \(\text{sech} u,\) is defined as \(\text{sech} u = \frac{2}{e^u + e^{-u}}.\)
By relating it to other hyperbolic functions, such as \(\tanh u = \frac{\sinh u}{\romanchar{10}\cosh u},\)
we were able to perform the differentiation accurately.
Utilizing these relationships streamlines the process, making complex differentiation problems more approachable.
Differentiation, in this context, not only provides the rate of change but also deepens our understanding of the behavior and properties of the Gudermannian and hyperbolic functions.