Problem 17 - Essential Calculus

Question

Find the derivative of the function. \(f(x)=\ln (\sinh x)\)

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = \ln (\sinh x)\) is \(\frac{\cosh x} {\sinh x}\).

Step 1: Identify the Inner Function

In the function f(x)=\(\ln (\sinh x)\), the inner function is \(\sinh x\). From the definition of hyperbolic functions, the derivative of \(\sinh x\) is \(\cosh x\). So \(v'(x) = \cosh x\).

Step 2: Identify the Outer Function and Derive it

The outer function in f(x)=\ln(sinh(x)) is \(\ln u\) where \(u = \sinh x\). The derivative of \(\ln u\) is \(\frac{1}{u}\). But as u is a function of x, we apply chain rule, so the derivative is \(\frac{u'(x)}{u(x)}\). In this case \(u'(x) = \cosh x\) and \(u(x) = \sinh x\).

Step 3: Apply Chain Rule

By applying the chain rule, we get the derivative of the function as the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function. Therefore, the derivative f'(x) = \(\frac{\cosh x} {\sinh x}\).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the trigonometric functions but for a hyperbola instead of a circle. They are important in various areas of mathematics and physics. In this context, the main functions we are interested in are the hyperbolic sine and hyperbolic cosine functions:

• The hyperbolic sine function, denoted as \( \sinh(x) \), is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).
• The hyperbolic cosine function, denoted as \( \cosh(x) \), is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).

A fascinating property of these functions is their relationship through derivatives. For example, the derivative of \( \sinh(x) \) is \( \cosh(x) \). This mirrors the behavior of regular sine and cosine functions in trigonometry but with hyperbolic partners instead. Understanding these derivatives is crucial for solving problems involving hyperbolic functions, like finding the derivative of \( \ln(\sinh(x)) \). In our exercise, the derivative of \( \sinh x \) directly plays a role when applying the chain rule.

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivative of a composite function. If you have two functions \( f(g(x)) \), the chain rule states that the derivative is \( f'(g(x)) \cdot g'(x) \). This means you take the derivative of the outer function evaluated at the inner function, then multiply it by the derivative of the inner function.

In the exercise, we start with the function \( f(x) = \ln(\sinh(x)) \). Here, the inner function is \( \sinh(x) \), while the outer function is \( \ln(u) \), where \( u = \sinh(x) \). By the chain rule, we calculate the derivative of \( \ln(u) \) as \( \frac{1}{u} \), then multiply by \( u'(x) = \cosh(x) \).

• The first step is deriving the inner function: \( \sinh(x) \rightarrow \cosh(x) \).
• Next, the derivative of the outer function \( \ln(\sinh x) \) becomes \( \frac{1}{\sinh(x)} \).
• Finally, combine using the chain rule: \( f'(x) = \frac{\cosh(x)}{\sinh(x)} \).

Thus, the chain rule enables us to systematically approach the differentiation of complex functions.

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique, especially useful when dealing with functions that are products, quotients, or powers. The idea is to take the natural logarithm of both sides of the equation \( y = f(x) \), differentiate implicitly with respect to \( x \), and then solve for \( y' \).

In our problem, logarithmic differentiation helps when differentiating the natural logarithm function \( \ln(\sinh(x)) \).

• The main advantage is simplifying complex expressions into manageable components.
• When differentiating \( \ln(\sinh(x)) \), we directly use the form \( \frac{d}{dx} [\ln(u)] = \frac{1}{u} \cdot u' \).

This approach makes it clear how each part of the function contributes to the final derivative. It's essential in cases where functions are deeply nested or complicated, allowing simplification of the differentiation process. In this exercise, applying this technique effectively through the chain rule gives a clear path to finding the derivative \( \frac{\cosh(x)}{\sinh(x)} \).