91Ó°ÊÓ

The inverse hyperbolic cotangent, \(\operatorname{coth^{-1} z}\), transforms the hyperbolic cotangent function to its inverse. To start with, it's essential to understand that the hyperbolic cotangent, \(\text{coth} x\), is defined as the ratio of the hyperbolic cosine to the hyperbolic sine: \(\text{coth} x = \frac{\text{cosh} x}{\text{sinh} x}\).
For deriving the logarithmic expression for \(\operatorname{coth^{-1} z}\), we express hyperbolic functions in their exponential forms. By substituting \(u = \operatorname{coth^{-1} z}\), it implies that \(z = \text{coth} u\). Therefore, \(z = \frac{e^u + e^{-u}}{e^u - e^{-u}}\).
Cross-multiplying, we get \(z(e^u - e^{-u}) = e^u + e^{-u}\). Simplifying further, we arrive at \(e^{2u} = \frac{z + 1}{z - 1}\). Upon taking the natural logarithm of both sides, we obtain \(2u = \text{ln} \frac{z+1}{z-1}\). Hence, \(u = \frac{1}{2} \text{ln} \frac{z+1}{z-1}\), leading to the final formula: \(\operatorname{coth^{-1} z} = \frac{1}{2} \text{ln} \frac{z+1}{z-1}\).

The inverse hyperbolic secant, \(\operatorname{sech^{-1} z}\), deals with the transformation of the hyperbolic secant function to its inverse. The hyperbolic secant, \(\text{sech} x\), is defined as \(\text{sech} x = \frac{2 e^{-x}}{e^{-2x} + 1}\).
By starting with \(u = \operatorname{sech^{-1} z}\), implying \(z = \text{sech} u\), we can write \(z = \frac{2}{e^u + e^{-u}}\). Solving for \(e^u\), we have \(z (e^u + e^{-u}) = 2\), which simplifies to the quadratic expression \(z e^u + z / e^u = 2\).
By letting \(v = e^u\), we substitute to form \(z v + \frac{z}{v} = 2\), resulting in the equation \(z v^2 - 2v + z = 0\).
Solving this quadratic equation using the quadratic formula, we obtain: \(v = \frac{1 + \sqrt{1 - z^2}}{z}\). Thus, the logarithmic expression for the inverse hyperbolic secant is: \(\operatorname{sech^{-1} z} = \log \left( \frac{1 + \sqrt{1 - z^2}}{z} \right)\).

Logarithmic expressions are mathematical formulations that use logarithms to simplify or solve equations involving exponentiation. To convert an exponential form into a logarithmic form, recall the basic logarithm definition \(y = \text{log}_b x\) if and only if \(b^y = x\).
In hyperbolic functions:

• The natural logarithm (log base e) is often used, as in functions involving exponential terms with base e (Euler's number).
• In the context of inverse hyperbolic functions, logarithmic expressions are derived by transforming exponential representations.
• For \(\operatorname{coth^{-1} z}\), converting \(e^u\) terms into logarithmic form simplifies the inversion process.
• Similarly, for \(\operatorname{sech^{-1} z}\), turning quadratic forms into logarithmic expressions helps solve for the variable more efficiently.

Understanding and familiarity with logarithms, exponent multiplication, and roots are crucial for deriving these inverse hyperbolic and other transcendental functions into manageable forms. Appreciating how these functions translate into logarithmic expressions will aid in their application and further mathematical studies.