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 \( D_x y \) is \( \frac{1}{\cosh^{-1}(x)\sqrt{x^2 - 1}} \).

Step 1: Understanding the Problem

We need to find the derivative of the function \( y = \ln(\cosh^{-1}(x)) \) with respect to \( x \). This involves understanding how to differentiate a composition of functions.

Step 2: Apply the Chain Rule

The function \( y \) is a composition of \( \ln(u) \) and \( u = \cosh^{-1}(x) \). According to the chain rule, the derivative is calculated as \( \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx} \). So, first, identify \( u = \cosh^{-1}(x) \) and determine \( \frac{du}{dx} \).

Step 3: Derivative of Inverse Hyperbolic Cosine

The derivative of \( \cosh^{-1}(x) \) is \( \frac{1}{\sqrt{x^2 - 1}} \). This is a known derivative for inverse hyperbolic functions.

Step 4: Differentiate

Using the chain rule and the derivative from the last step, we apply it to find \( \frac{dy}{dx} \):\[\frac{dy}{dx} = \frac{1}{\cosh^{-1}(x)} \cdot \frac{1}{\sqrt{x^2 - 1}}.\]

Step 5: Simplify the Expression

The derivative of the given function is itself a product of fractions. The simplified expression for \( \frac{dy}{dx} \) remains as presented:\[\frac{dy}{dx} = \frac{1}{\cosh^{-1}(x)\sqrt{x^2 - 1}}.\]

Key Concepts

Chain Rule

When faced with a composition of functions, like in our function \( y = \ln(\cosh^{-1}(x)) \), we need to use the chain rule to differentiate it. The chain rule is a fundamental technique in calculus used to find the derivative of a composite function, which is a function created from two or more functions. In essence, the rule states:

• If we have a function \( y = f(g(x)) \), the derivative \( \frac{dy}{dx} \) is found by multiplying the derivative of the outer function \( f \) evaluated at the inner function \( g(x) \), by the derivative of the inner function \( g(x) \) itself, or \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

In our exercise, \( f(u) = \ln(u) \) and \( g(x) = \cosh^{-1}(x) \). Thus, differentiating \( y = \ln(\cosh^{-1}(x)) \) requires finding:

• \( \frac{d}{du} (\ln(u)) = \frac{1}{u} \) for the outer function.
• \( \frac{du}{dx} \text{ (where } u = \cosh^{-1}(x)) \) for the inner function, which we will explore next.

Inverse Hyperbolic Functions

Inverse hyperbolic functions reverse the effect of hyperbolic functions, much like inverse trigonometric functions. They are particularly helpful in many calculus and physics problems. In this problem, we have the inverse hyperbolic cosine, \( \cosh^{-1}(x) \). Alongside knowing the basic hyperbolic functions, understanding their inverses and derivatives is critical:

• The function \( \cosh(x) \) maps the real line to values \( [1, \infty) \).
• The inverse, \( \cosh^{-1}(x) \), maps values in the interval \( [1, \infty) \) back to real numbers.

The derivative of \( \cosh^{-1}(x) \) is essential:

• \( \frac{d}{dx} (\cosh^{-1}(x)) = \frac{1}{\sqrt{x^2 - 1}} \). This derivative arises from its natural definition, capturing how the rate of change behaves with respect to \( x \), provided \( x > 1 \).

This result allows us to proceed with finding the derivative of our composite function.

Derivative Simplification

After applying the chain rule and finding the derivatives of each function involved, it's crucial to simplify the expression at the end. Derivative simplification can make complex results more understandable and more useful for further applications.In the solution to our exercise, we arrive at the expression:

• \( \frac{dy}{dx} = \frac{1}{\cosh^{-1}(x)} \cdot \frac{1}{\sqrt{x^2 - 1}} \).

Simplification means organizing this product of fractions:

• The final step was to write \( \frac{dy}{dx} \) as \( \frac{1}{\cosh^{-1}(x) \sqrt{x^2 - 1}} \).
• While it looks simple, each part plays a significant role. Being concise helps ensure accurate communication of complex mathematics.

Remember, simplifying doesn't always mean making it shorter; it means making it easier to understand and applying further in calculations.