Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\ln \left(\cosh ^{-1} x\right) $$

Step-by-Step Solution

Answer

The derivative is \( \frac{1}{\cosh^{-1}(x) \cdot \sqrt{x^2 - 1}} \).

Step 1: Understand the Problem

We need to find the derivative of the function \( y = \ln(\cosh^{-1} x) \) with respect to \( x \). This involves differentiating a natural logarithm function composed with an inverse hyperbolic cosine function.

Step 2: Differentiate the Outer Function

The outer function is the natural logarithm \( \ln(u) \), where \( u = \cosh^{-1}(x) \). The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). Thus, \( \frac{d}{du} \ln(u) = \frac{1}{u} \).

Step 3: Differentiate the Inner Function

The inner function is \( u = \cosh^{-1}(x) \). The derivative of \( \cosh^{-1}(x) \) with respect to \( x \) is \( \frac{1}{\sqrt{x^2-1}} \). Thus, \( \frac{du}{dx} = \frac{1}{\sqrt{x^2-1}} \).

Step 4: Apply the Chain Rule

Using the chain rule for differentiation, \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Substitute the derivatives found in Steps 2 and 3: \( \frac{dy}{dx} = \frac{1}{\cosh^{-1}(x)} \cdot \frac{1}{\sqrt{x^2 - 1}} \).

Step 5: Simplify the Expression

Combine the expressions to get the final form of the derivative: \( \frac{dy}{dx} = \frac{1}{\cosh^{-1}(x) \cdot \sqrt{x^2 - 1}} \).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions that arise from a hyperbola, much like trigonometric functions arise from a circle. The hyperbolic cosine function, denoted as \( \cosh(x) \), is one such function and is defined by the formula:

• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

The inverse hyperbolic cosine, \( \cosh^{-1}(x) \), is the function that "undoes" \( \cosh(x) \). This function tells us the angle whose hyperbolic cosine is \( x \). It's similar to how the inverse cosine function works with trigonometric circles. The formula for \( \cosh^{-1}(x) \) is:

• \( \cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1}) \)

This function is primarily used in calculus to deal with integrals and derivatives and can be somewhat complex due to its connection with exponential functions. Remember that the domain for \( \cosh^{-1}(x) \) is \( x \geq 1 \).
Understanding these functions is crucial as they play a significant role in advanced mathematical contexts and appear often in engineering and physics.

Derivative

A derivative is a measure of how a function changes as its input changes. It tells us the rate of change or the slope of the function at any given point. In this exercise, we are finding the derivative of \( y = \ln(\cosh^{-1}(x)) \) with respect to \( x \). This involves the concept of taking the derivative first of the natural logarithm, and then of the inverse hyperbolic cosine.
To start, we know the derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). This is an important rule when differentiating composite functions. Next, we need the derivative of the inverse hyperbolic cosine function, \( \cosh^{-1}(x) \), which is \( \frac{1}{\sqrt{x^2-1}} \). This requires integration knowledge and understanding of hyperbolic identities.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. It's an essential function in calculus because of its simple derivative and integral properties. In the exercise, you're working with \( \ln(u) \), where \( u = \cosh^{-1}(x) \).
The derivative of the natural logarithm function has a straightforward rule: \( \frac{d}{du} \ln(u) = \frac{1}{u} \). This rule simplifies the differentiation process significantly and shows why the natural logarithm is often used in calculus problems.
When differentiating composite functions that involve \( \ln(\text{something}) \), you need to be vigilant of the chain rule, as it ensures the proper sequence of differentiation steps.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate compositions of functions. It states that if a function \( y \) depends on \( u \), which in turn depends on \( x \), then the derivative of \( y \) with respect to \( x \) is:

• \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

This rule is especially useful when working with nested functions, such as \( y = \ln(\cosh^{-1}(x)) \). Here, you differentiate the outer function, \( \ln(u) \), with respect to \( u \), and then multiply it by the derivative of the inner function, \( \cosh^{-1}(x) \), with respect to \( x \).
Understanding how to apply the chain rule is essential for tackling complex calculus problems. It breaks down complicated derivative tasks into manageable parts.