Problem 20 - Calculus

Question

Find the derivatives of the functions \(9-18:\) From \(\tanh x=\frac{3}{5}\) find \(\operatorname{sech} x, \cosh x, \sinh x, \operatorname{coth} x, \operatorname{csch} x\)

Step-by-Step Solution

Answer

\(\operatorname{sech} x = \frac{4}{5}\), \(\cosh x = \frac{5}{4}\), \(\sinh x = \frac{3}{4}\), \(\operatorname{coth} x = \frac{5}{3}\), \(\operatorname{csch} x = \frac{4}{3}\).

Step 1: Recall the relationship between hyperbolic functions

The given function is \(\tanh x = \frac{3}{5}\). Recall that \(\tanh x = \frac{\sinh x}{\cosh x}\) which gives us the equation: \(\frac{\sinh x}{\cosh x} = \frac{3}{5}\).

Step 2: Use the identity \(\cosh^2 x - \sinh^2 x = 1\)

From the identity \(\cosh^2 x - \sinh^2 x = 1\), we will solve for \(\cosh x\) and \(\sinh x\) by expressing them in terms of one another.

Step 3: Solve for \(\sinh x\)

Using \(\sinh x = \frac{3}{5} \cosh x\) from step 1, substitute \(\sinh x\) in the identity: \[\cosh^2 x - \left(\frac{3}{5} \cosh x\right)^2 = 1\] Simplifying gives \[\cosh^2 x - \frac{9}{25} \cosh^2 x = 1\] which further simplifies to \[\frac{16}{25} \cosh^2 x = 1\].

Step 4: Calculate \(\cosh x\)

Solving \(\frac{16}{25} \cosh^2 x = 1\) gives \(\cosh^2 x = \frac{25}{16}\), therefore \(\cosh x = \frac{5}{4}\) (choosing the positive value since hyperbolic functions are defined in all real numbers).

Step 5: Calculate \(\sinh x\)

Using \(\sinh x = \frac{3}{5} \cosh x\), substitute \(\cosh x = \frac{5}{4}\) to find \[\sinh x = \frac{3}{5} \times \frac{5}{4} = \frac{3}{4}\].

Step 6: Calculate \(\operatorname{sech} x\)

\(\operatorname{sech} x = \frac{1}{\cosh x}\). From step 4, \(\cosh x = \frac{5}{4}\), so \[\operatorname{sech} x = \frac{1}{\frac{5}{4}} = \frac{4}{5}\].

Step 7: Calculate \(\operatorname{coth} x\)

\(\operatorname{coth} x = \frac{\cosh x}{\sinh x}\). Substitute \(\cosh x = \frac{5}{4}\) and \(\sinh x = \frac{3}{4}\) to get \[\operatorname{coth} x = \frac{\frac{5}{4}}{\frac{3}{4}} = \frac{5}{3}\].

Step 8: Calculate \(\operatorname{csch} x\)

\(\operatorname{csch} x = \frac{1}{\sinh x}\). From step 5, \(\sinh x = \frac{3}{4}\), so \[\operatorname{csch} x = \frac{1}{\frac{3}{4}} = \frac{4}{3}\].

Key Concepts

Understanding Derivatives in Hyperbolic Functions

Derivatives help us understand how functions change. Hyperbolic functions, like sine and cosine, have their own versions known as sinh and cosh. They're essential in calculus and provide a framework for solving equations involving rapid changes.
The derivative of the hyperbolic functions can be expressed similar to those of trigonometric functions. For example,

• The derivative of \(\sinh x\) is \(\cosh x\).
• The derivative of \(\cosh x\) is \(\sinh x\).
• The derivative of \(\tanh x\) is \(\operatorname{sech}^2 x\).

That's because these derivatives resemble the chain of change between sine and cosine. Of course, understanding these derivatives helps us predict how each of these functions will behave and allows us to compute answers relating to their changes effectively.

Unpacking the Tanh Function

The hyperbolic tangent function, or \(\tanh\), is one of the key concepts in hyperbolic functions. It's defined as the ratio of \(\sinh x\) to \(\cosh x\). Mathematically, it is written as: \[\tanh x = \frac{\sinh x}{\cosh x}\].
This function is significant because it ranges between -1 and 1, and represents the shape of a hyperbola in a way that's useful for mathematical applications.
Some important characteristics of \(\tanh x\) are:

• \(\tanh x\) is an odd function, which means \(\tanh (-x) = -\tanh x\).
• Its value approaches 1 as \(x\) goes to infinity, and -1 as \(x\) approaches negative infinity.
• Its derivative is \(\operatorname{sech}^2 x\), a crucial aspect in calculus for determining slopes.

This function is widely used in physics and engineering, particularly in real-world applications like hyperbolic geometry and relativistic computations.

Cosh and Sinh Identities Explained

Understanding the identities that relate \(\cosh x\) and \(\sinh x\) is crucial.
The most fundamental identity is: \[\cosh^2 x - \sinh^2 x = 1\].
This equation mirrors the Pythagorean identity of trigonometric functions but applies to hyperbolic functions.
Let's review other crucial attributes:

• \(\cosh x = \frac{e^x + e^{-x}}{2}\)
• \(\sinh x = \frac{e^x - e^{-x}}{2}\)
• These functions show the connection between the exponential function and the shape of hyperbolas.

Understanding these identities allows you to solve various problems involving hyperbolic functions by simplifying complex equations into solvable forms. By recognizing these patterns, complex ideas become manageable and approachable.

Hyperbolic Function Identities

Hyperbolic functions have identities similar to those in trigonometry. They describe relationships between functions like \(\tanh x\), \(\cosh x\), and \(\sinh x\).
Some key hyperbolic identities include:

• \(\operatorname{sech} x = \frac{1}{\cosh x}\) - helps in reversing \(\cosh\).
• \(\operatorname{coth} x = \frac{\cosh x}{\sinh x}\) - represents reciprocal of \(\tanh x\).
• \(\operatorname{csch} x = \frac{1}{\sinh x}\) - finds reciprocal value for \(\sinh x\).

Recognizing these identities can simplify complex expressions or verify solutions, as they relate to constants already known from step-by-step problem-solving processes.
By leveraging these hyperbolic identities, you can solve expressions more efficiently, understanding both trigonometric and exponential aspects in applications from mathematics to physics.