Problem 28 - Essential Calculus Early Transcendentals

Question

Find the derivative. Simplify where possible. possible. $$ g(x)=\cosh (\ln x) $$

Step-by-Step Solution

Answer

The derivative of \( g(x) = \cosh(\ln x) \) is \( g'(x) = \frac{\sinh(\ln x)}{x} \).

Step 1: Identify the Composition

To differentiate the function \( g(x) = \cosh(\ln x) \), notice that it is a composition of two functions: the hyperbolic cosine function \( \cosh(u) \) and \( u = \ln x \). We will need to use the chain rule to differentiate this composition.

Step 2: Differentiate Outer Function

Differentiate the outer function \( \cosh(u) \) in terms of \( u \). The derivative is \( \sinh(u) \). So, \( \frac{d}{du}[\cosh(u)] = \sinh(u) \).

Step 3: Differentiate Inner Function

Differentiate the inner function \( u = \ln x \) in terms of \( x \). The derivative is \( \frac{1}{x} \). Thus, \( \frac{d}{dx}[\ln x] = \frac{1}{x} \).

Step 4: Apply the Chain Rule

Use the chain rule to combine these derivatives: \( \frac{d}{dx}[\cosh(\ln x)] = \sinh(\ln x) \cdot \frac{1}{x} \).

Step 5: Simplify the Expression

Simplify the expression for the derivative: \( g'(x) = \frac{\sinh(\ln x)}{x} \). This is already simplified, as there are no further simplifications possible.

Key Concepts

Chain Rule

The chain rule is an essential technique in calculus, especially when you need to differentiate a composite function. If you have a function that is applied to another function, you need this rule to split the task into more manageable pieces. Imagine the function as two layers: an outer layer and an inner layer. You start by differentiating the outer layer while treating the inner one as a fixed point. Then, you differentiate the inner layer. Finally, you multiply these derivatives together.For example, in the function \( g(x) = \cosh(\ln x) \), the outer function is \( \cosh(u) \) and the inner function is \( u = \ln x \). First, differentiate \( \cosh(u) \) to get \( \sinh(u) \), then differentiate \( \ln x \) to get \( \frac{1}{x} \). By the chain rule, we combine them: \[ g'(x) = \sinh(\ln x) \cdot \frac{1}{x} \].

• Outer Function: Differentiate \( \cosh(u) \) to get \( \sinh(u) \).
• Inner Function: Differentiate \( \ln x \) to get \( \frac{1}{x} \).
• Final Combination: Multiply the derivatives to get \( \sinh(\ln x) \cdot \frac{1}{x} \).

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola instead of a circle. They have unique properties that make them useful in various branches of mathematics and engineering. Useful hyperbolic functions include hyperbolic sine \( \sinh(x) \), hyperbolic cosine \( \cosh(x) \), as well as hyperbolic tangent \( \tanh(x) \). When differentiating these functions, you follow specific rules just as in trigonometric differentiation.Interestingly, the derivative of \( \cosh(x) \) is \( \sinh(x) \), and vice versa. In our problem, \( g(x) = \cosh(\ln x) \), so its derivative involves \( \sinh(\ln x) \). Hyperbolic functions always relate back to exponential functions; for instance, \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).

• Derivative Insight: For \( \cosh(x) \), its derivative is \( \sinh(x) \).
• Inverse Relationship: If \( \frac{d}{dx}[\cosh(x)] = \sinh(x) \), then \( \frac{d}{dx}[\sinh(x)] = \cosh(x) \).
• Exponential Form: \( \cosh(x) \) links to exponentials: \( \frac{e^x + e^{-x}}{2} \).

Logarithmic Differentiation

Logarithmic differentiation is a technique to find the derivative of functions that contain logarithms or those that can be simplified using logarithms. It's particularly helpful for complicated products or quotients of functions. The key idea is to take the natural logarithm of the function and differentiate it using simple differentiation rules.In our example, the inner function \( \ln x \) is simple enough, and its derivative is straightforward, \( \frac{d}{dx}[\ln x] = \frac{1}{x} \). Often, this method simplifies dealing with powers or products, but here it directly relates to dealing with the composite function in the chain rule.

• Simple Derivative: For \( \ln x \), \( \frac{d}{dx}[\ln x] = \frac{1}{x} \).
• Complexity Reduction: Great for simplifying products or quotients.
• Chain Rule: Here, it integrates seamlessly into the chain rule application.