Problem 27 - Calculus of a Single Variable

Question

In Exercises 23–32, 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 \( f'(x) = \coth(x) \).

Step 1: Identify the Outer Function and the Inner Function

Recognize that the function \( f(x) = \ln (\sinh x) \) is a composition of two functions. Here, the outer function is \( \ln(u) \) and the inner function is \( u = \sinh(x) \). Derivation of a function composition requires the chain rule.

Step 2: Apply the Chain Rule

The derivative of a composition of functions can be calculated using the chain rule. The chain rule is formally written as: \( (f(g(x)))' = f'(g(x)) * g'(x) \) where \( f(g(x)) \) is our function, \( f'(g(x)) \) is the derivative of the outer function evaluated at the inner function, and \( g'(x) \) is the derivative of the inner function. Apply this to our function. We begin by finding the derivative of the outer function, which is \( f'(u) = 1/u \). Then substitute \( u = \sinh(x) \), yielding \( f'(u) = 1/\sinh(x) \).

Step 3: Find the Derivative of the Inner Function

Find the derivative of \( u = \sinh(x) \). The derivative of hyperbolic sine is hyperbolic cosine, i.e., \( \sinh'(x) = \cosh(x) \).

Step 4: Combine using the Chain Rule

According to the chain rule, the derivative of \( f(x) \) will be the derivative of the outer function multiplied by the derivative of the inner function. This gives \( f'(x) = 1/\sinh(x) * \cosh(x) \).

Step 5: Simplify the Result

In hyperbolic functions, the relationship \( \cosh^2(x) - \sinh^2(x) = 1 \) holds. Using this, we can simplify the result further as \( f'(x) = \cosh(x)/\sinh(x) = \coth(x) \), which is the derivative of the given function.

Key Concepts

Chain Rule

The chain rule is a powerful tool for finding the derivative of a composition of functions. It tells us how to take derivatives of nested functions, which is when one function is inside another. This rule can be formally expressed as:

• \((f(g(x)))' = f'(g(x)) \times g'(x)\)

To use the chain rule, you need to identify the outer and inner functions. In our problem, the outer function is \(\ln(u)\) and the inner function is \(u = \sinh(x)\). You first find the derivative of the outer function, replacing the inside part with the inner function. Next, you multiply by the derivative of the inner function. This combination steps together, like a chain, producing the required derivative. This approach simplifies even more complex problems into manageable pieces.

Hyperbolic Functions

Hyperbolic functions are analogs of regular trigonometric functions but relate to hyperbolas, not circles. Some common hyperbolic functions are \(\sinh(x)\) and \(\cosh(x)\). These functions may seem intimidating, but they have straightforward derivatives:

• The derivative of \(\sinh(x)\) is \(\cosh(x)\).
• The derivative of \(\cosh(x)\) is \(\sinh(x)\).

These relationships are similar to the derivatives of sine and cosine in trigonometry but with some differences in sign. Hyperbolic identities, like \(\cosh^2(x) - \sinh^2(x) = 1\), are useful for simplifying expressions. Knowing these derivatives and identities is essential when working with equations that involve hyperbolic functions.

Composition of Functions

A composition of functions occurs when one function is inside another. In mathematical terms, this can be written as \(f(g(x))\). Understanding compositions is crucial for applying the chain rule. It's about nesting functions properly.

• Identify which part is the outer function \(f(x)\).
• Identify the inner function \(g(x)\).

For example, with \(f(x) = \ln(\sinh(x))\), "\(\ln\)" is the outer function and "\(\sinh(x)\)" is the inner function. Breaking the function into these components allows us to apply the chain rule effortlessly. This method of peeling layers one at a time simplifies derivatives that would otherwise be complex and hard to manage.