Problem 48 - Essential Calculus Early Transcendentals

Question

\(47-48=\) Differentiate \(f\) and find the domain of \(f\) $$f(x)=\ln \ln \ln x$$

Step-by-Step Solution

Answer

Differentiate: \(\frac{1}{\ln(\ln(x))\ln(x)x}\); Domain: \(x>e\).

Step 1: Understand Function Composition

The function given is a composition of logarithmic functions: \( f(x) = \ln(\ln(\ln(x))) \). This means we have three nested natural logarithms, each requiring the input to be greater than 0.

Step 2: Determine Innermost Domain Requirement

The innermost function is \( \ln(x) \), which requires \( x > 0 \). In addition to this, for \( \ln(\ln(x)) \) to be defined, \( \ln(x) > 0 \) must hold, which implies \( x > e^0 = 1 \).

Step 3: Determine Middle Function Domain Requirement

Next, we need to ensure \( \ln(\ln(x)) \) is greater than 0 for \( \ln(\ln(\ln(x))) \) to be defined. This results in \( \ln(\ln(x)) > 0 \), leading to \( \ln(x) > e^0 = 1 \), which simplifies to \( x > e^1 = e \).

Step 4: Combine Domain Restrictions

The most restrictive requirement comes from needing \( x > e \). Therefore, the domain of \( f(x) \) is \( x > e \).

Step 5: Differentiate the function using the Chain Rule

Differentiate \( f(x) = \ln(\ln(\ln(x))) \) using the chain rule. Let \( u = \ln(x) \), \( v = \ln(u) \), and \( w = \ln(v) \). Then \( f'(x) = \frac{1}{w} \cdot \frac{1}{v} \cdot \frac{1}{u} \cdot \frac{1}{x} = \frac{1}{\ln(\ln(x)) \cdot \ln(x) \cdot x}.\)

Key Concepts

Chain Rule

The chain rule is a vital concept in calculus used to differentiate composite functions. It allows us to take the derivative of a function that is nested within another. Imagine a sequence of functions, each built upon the previous one.
For the function given in the problem, we have three layers of logarithms: \( f(x) = \ln(\ln(\ln(x))) \). To differentiate this, we apply the chain rule repeatedly.
First, define the innermost component as \( u = \ln(x) \), the second layer as \( v = \ln(u) \), and the overall function as \( w = \ln(v) \). The derivative of \( w \) with respect to \( x \) involves differentiating each layer:

• The derivative of \( w \) with respect to \( v \) is \( 1/v \).
• The derivative of \( v \) with respect to \( u \) is \( 1/u \).
• The derivative of \( u \) with respect to \( x \) is \( 1/x \).

Therefore, by multiplying these together, we apply the chain rule and find \\[f'(x) = \frac{1}{\ln(\ln(x)) \cdot \ln(x) \cdot x}.\] This ensures a comprehensive application of the chain rule over multiple layers.

Logarithmic Function

Logarithmic functions are the inverse of exponential functions and are crucial for various applications in mathematics and science. For our function, \( f(x) = \ln(\ln(\ln(x))) \), each \( \ln \) denotes the natural logarithm, which is the logarithm to the base \( e \).
When you see multiple logarithms nested within each other, like in this problem, it means each log operates on the result of the previous one.
The properties of logarithmic functions impact how we structure our calculations:

• The natural logarithm \( \ln(x) \) is only defined for \( x > 0 \).
• For \( \ln \ln(x) \) to exist, \( \ln(x) \) must itself be positive.
• As we progress to \( \ln \ln \ln(x) \), the same rule continues to propagate through each layer, affecting the domain.

Understanding these concepts allows us to tackle, simplify, and differentiate nested logarithmic functions correctly.

Function Domain

Determining the domain of a function is vital to understanding where the function is valid and can be computed. The function \( f(x) = \ln(\ln(\ln(x))) \) consists of three nested logarithms, each imposing specific requirements on the input \( x \).
Since our function is composed of natural logarithms:

• The innermost logarithm \( \ln(x) \) requires \( x > 0 \).
• Next, \( \ln(\ln(x)) \) requires that \( \ln(x) > 0 \), leading to \( x > 1 \).
• Finally, \( \ln(\ln(\ln(x))) \) further requires \( \ln(\ln(x)) > 0 \), implying \( x > e \) due to \( \ln(x) > 1 \).

The strictest condition dictates the domain requirement: \( x > e \). Thus, understanding the interplay of each logarithm layer allows us to accurately define the domain for this composite function.