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\(27-30\) Differentiate \(f\) and find the domain of \(f\) \(f(x)=\ln \ln \ln x\)

Short Answer

Expert verified
The derivative is \( f'(x) = \frac{1}{x \ln x \ln(\ln x)} \) and the domain is \( x > e^e \).

Step by step solution

01

Differentiate the Outer Function

The function we have is a composition of three natural logarithmic functions: \( f(x) = \ln(\ln(\ln x)) \). To differentiate this, we will use the chain rule. Start by differentiating the outermost function, which is \( \ln u \) (where \( u = \ln(\ln x) \)). The derivative of \( \ln u \) is \( \frac{1}{u} \), so the derivative for the outer layer is \( \frac{1}{\ln(\ln x)} \).
02

Differentiate the Middle Function

Now, differentiate the middle function \( u = \ln(\ln x) \). For this, we have \( u = \ln v \) (where \( v = \ln x \)). The derivative of \( \ln v \) is \( \frac{1}{v} \), so the derivative is \( \frac{1}{\ln x} \).
03

Differentiate the Innermost Function

Finally, differentiate the innermost function \( v = \ln x \). The derivative of \( \ln x \) is \( \frac{1}{x} \). This gives us the derivative of the innermost layer.
04

Apply the Chain Rule

To find the derivative of the entire function, we use the chain rule to multiply the derivatives from each step:\[ f'(x) = \frac{1}{\ln(\ln x)} \cdot \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x \ln(\ln x)} \].
05

Determine the Domain of \( f(x) \)

The function \( f(x) = \ln(\ln(\ln x)) \) requires each argument of the natural log to be positive. Firstly, \( x > 1 \) so that \( \ln x > 0 \); then \( \ln x > 1 \) so that \( \ln(\ln x) > 0 \); and finally, since \( \ln(\ln x) > 0 \), \( \ln(\ln x) > 1 \). Solving \( \ln x > 1 \) implies \( x > e \) and further for \( \ln(\ln x) > 0 \), solving gives \( x > e^e \). Hence, the domain of \( f \) is \( x > e^e \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Logarithm
The natural logarithm, often denoted as \( \ln x \), is a fundamental concept in calculus. It is the logarithm to the base \( e \), where \( e \approx 2.71828 \) is an important mathematical constant. This log function is defined only for positive values, meaning \( x > 0 \). The natural log of a number \( x \) gives the power to which \( e \) must be raised to obtain \( x \).A few essential properties of the natural logarithm include:
  • \( \ln 1 = 0 \) because \( e^0 = 1 \).
  • \( \ln e = 1 \) as \( e^1 = e \).
  • For any positive \( a \) and \( b \), \( \ln(ab) = \ln a + \ln b \).
  • \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \).
These properties are useful when simplifying expressions involving natural logarithms. In our task, since we have multiple logs, understanding each layer of the natural log is critical for correctly applying differentiation techniques, such as the chain rule.
Derivative of Composite Functions
Differentiating composite functions is a key skill in calculus, particularly when they involve functions nested within each other, or compositions. To tackle the derivative of such functions efficiently, we utilize the chain rule.

The Chain Rule

The chain rule states that to find the derivative of a composite function \( f(g(x)) \), you first differentiate the outer function \( f \), and then multiply it by the derivative of the inner function \( g \). In formula terms, it’s expressed as:\[(f(g(x)))' = f'(g(x)) \cdot g'(x)\]In the exercise, the function is \( f(x) = \ln(\ln(\ln x)) \). This is composed of three nested functions. Following the chain rule:
  • First, differentiate \( \ln(\ln x) \) with respect to \( \ln(\ln x) \), giving \( \frac{1}{\ln(\ln x)} \).
  • Next, take \( \ln x \) inside \( \ln(\ln x) \), derivative is \( \frac{1}{\ln x} \).
  • Finally, differentiate \( \ln x \) to get \( \frac{1}{x} \).
The overall derivative combines these, resulting in \( \frac{1}{x \ln x \ln(\ln x)} \), effectively layering the derivatives through multiplication.
Domain of a Function
The domain of a function refers to all the possible inputs (or \( x \) values) for which the function is defined and produces real numbers. For functions involving natural logarithms, like in our example, it's key to ensure that the expression inside the logarithms remains positive, as the natural log is undefined for non-positive values.

Determining the Domain

For \( f(x) = \ln(\ln(\ln x)) \), we need to find values of \( x \) for which each \( \ln \) term is positive:
  • Firstly, \( \ln x > 0 \) implies \( x > 1 \).
  • Secondly, \( \ln(\ln x) > 0 \) implies \( \ln x > 1 \), which simplifies to \( x > e \).
  • Finally, \( \ln(\ln(\ln x)) > 0 \) implies \( \ln(\ln x) > 1 \), leading to solving \( \ln x > e \), ultimately meaning \( x > e^e \).
Thus, for the function to be defined, \( x \) must be greater than \( e^e \), making this the domain. These steps ensure calculations remain valid and the function produces real-valued results.

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