Problem 20 - Calculus Single Variable

Question

Calculate the derivative with respect to \(x\) of the given expression. \(\ln (\ln (x))\)

Step-by-Step Solution

Answer

\( \frac{1}{x \ln(x)} \)

Step 1: Identify Outer Function

To differentiate the expression \( \ln (\ln (x)) \), first recognize that the outer function is \( \ln(u) \) where \( u = \ln(x) \).

Step 2: Apply Chain Rule

The chain rule states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \). Here, \( f(u) = \ln(u) \), so its derivative is \( \frac{1}{u} \) and \( g(x) = \ln(x) \), so its derivative is \( \frac{1}{x} \).

Step 3: Compute Derivative of Outer Function

Differentiate the outer function \( f(u) = \ln(u) \) with respect to \( u \). The derivative is \( \frac{1}{u} \).

Step 4: Compute Derivative of Inner Function

Differentiate the inner function \( g(x) = \ln(x) \). The derivative is \( \frac{1}{x} \).

Step 5: Combine Using the Chain Rule

Combine the derivatives from Steps 3 and 4 using the chain rule: \( \frac{1}{\ln(x)} \cdot \frac{1}{x} \).

Step 6: Simplify the Expression

Thus, the derivative of the original expression \( \ln(\ln(x)) \) with respect to \( x \) is \( \frac{1}{x \ln(x)} \).

Key Concepts

Differentiation

Differentiation is a key concept in calculus that deals with finding the rate at which a function is changing at any given point. It is essentially the process of finding a derivative. The derivative, represented as \( f'(x) \) or \( \frac{df}{dx} \), describes how a function f changes as its input changes.

When we differentiate a function, we are looking for its slope or the instantaneous rate of change. For instance, if we have a function \( y = f(x) \), the derivative \( \frac{dy}{dx} \) tells us how much y would change when x changes by a small amount.

Differentiation rules, such as the product rule, quotient rule, and chain rule, provide specific methods to find derivatives of different types of functions. Understanding differentiation is vital as it lays the foundation for solving optimization problems, analyzing trends, and modeling real-world situations.

Composite Functions

A composite function is formed when one function is applied to the result of another function. It's written as \( (f \circ g)(x) = f(g(x)) \). In simpler terms, you perform the inside function first, then the outside function. For example, in the expression \( \ln(\ln(x)) \), \( \ln(x) \) is the inner function, and \( \ln(u) \) with \( u = \ln(x) \) is the outer function.

To differentiate composite functions, we use the chain rule. This is crucial because it allows us to handle functions that are not simple polynomials or basic trigonometric expressions. The chain rule states:

• Differentiating the outer function first, while keeping the inner function unchanged.
• Multiplying it by the derivative of the inner function.

Understanding composite functions and their derivatives helps in analyzing complex behaviors where functions depend on each other. It is widely used in mathematics, physics, and engineering to model real-life scenarios.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions and are represented as \( y = \log_b(x) \), where \( b \) is the base of the logarithm. A common example is the natural logarithm \( \ln(x) \), where the base is \( e \), an irrational constant approximately equal to 2.718.

Logarithmic functions are particularly valuable because they can simplify complex multiplication and division into addition and subtraction, making calculations more manageable. The derivative of \( \ln(x) \), for example, is \( \frac{1}{x} \), a fundamental result used in many calculus problems.

When dealing with nested logarithmic functions like \( \ln(\ln(x)) \), it's important to recognize their composite nature. By applying the chain rule, we can differentiate such expressions by first focusing on the outer logarithmic function and then the inner one. These functions are used extensively in growth models, signal processing, and in learning machine algorithms.