Problem 29 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\ln (\ln x)$$

Step-by-Step Solution

Answer

The derivative of \( y = \ln(\ln x) \) with respect to \( x \) is \( \frac{1}{x \ln x} \).

Step 1: Identify the Function

Our function is given as \( y = \ln(\ln x) \). Here, \(y\) is expressed in terms of \(x\), so we need to find \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\).

Step 2: Apply Chain Rule

Recognize that \(\ln(\ln x)\) is a composite function, where \( u = \ln x \), so \( y = \ln(u) \). We must use the chain rule which states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).

Step 3: Find Derivative of Outer Function

For \( y = \ln(u) \), the derivative with respect to \(u\) is \( \frac{1}{u} \). That is, \( \frac{dy}{du} = \frac{1}{u} \).

Step 4: Find Derivative of Inner Function

The inner function is \( u = \ln x \). The derivative of \( \ln x \) with respect to \( x \) is \( \frac{1}{x} \). Thus, \( \frac{du}{dx} = \frac{1}{x} \).

Step 5: Combine Derivatives Using Chain Rule

Apply the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Substitute the derivatives from steps 3 and 4: \( \frac{dy}{dx} = \frac{1}{u} \cdot \frac{1}{x} = \frac{1}{\ln x} \cdot \frac{1}{x} \).

Step 6: Simplify the Expression

The expression \( \frac{1}{\ln x} \cdot \frac{1}{x} \) simplifies to \( \frac{1}{x \ln x} \). Thus, \( \frac{dy}{dx} = \frac{1}{x \ln x} \).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus used for finding the derivative of composite functions. A composite function occurs when you have a function nested within another function, like a layered cake. In our exercise, the function \( y = \ln(\ln x) \) is a composite function because \( \ln x \) is nested within another natural logarithm function, \( \ln(u) \), where \( u = \ln x \).

The chain rule helps us differentiate such functions by breaking them down into their outer and inner components. For instance, if you have a composite function \( f(g(x)) \), the chain rule states that its derivative is the derivative of the outer function \( f \) evaluated at the inner function \( g \), multiplied by the derivative of the inner function \( g \). In simple terms, it's like peeling off the outer layer first, differentiating it, and then moving to the next inner layer.

In formulaic terms, for our function, this means:

• First, differentiate the outer function: Here, \( f(u) = \ln(u) \), with its derivative being \( \frac{1}{u} \).
• Next, differentiate the inner function: \( g(x) = \ln x \), with its derivative being \( \frac{1}{x} \).

Then, the chain rule tells us to multiply these derivatives: \( \frac{dy}{dx} = \frac{1}{u} \cdot \frac{1}{x} \), which eventually simplifies to the solution we found: \( \frac{1}{x \ln x} \).

Composite Functions

Composite functions occur frequently in calculus and involve several layers of functions wrapped around each other like an onion. In the problem given, \( y = \ln(\ln x) \), we encounter a scenario where one function is inserted into another.

To break it down, think of composite functions like a box within a box:

• The outer box (or function) is \( \ln(u) \), here dealing with the result of another function.
• The inner box is \( \ln x \), which outputs the result that the outer function will process.

This layered structure means when taking the derivative, we need to consider both functions separately before combining their derivatives using the chain rule.

Recognizing these layers is essential for correctly applying differentiation techniques, like the chain rule. When attacking problems involving composite functions, focus on identifying inner and outer functions to simplify the process of finding derivatives.

Logarithmic Differentiation

Logarithmic differentiation is a clever differentiation technique, especially useful when dealing with complex functions like logarithms and products of functions. It involves the logarithmic properties to simplify the differentiation process.

In our exercise, using logarithmic differentiation offers a structured manner to handle functions like \( y = \ln(\ln x) \). The main advantage is it leverages log properties to transform challenging derivatives into more manageable forms. For example, properties like \( \ln(a \times b) = \ln a + \ln b \) and \( \ln(a^b) = b \ln a \) can break down difficult expressions into easier components.

While our example does not fully utilize logarithmic differentiation since no extra complexity arises, it's important to highlight how it could be used for similar problems:

• First, take the natural log of both sides: Converts multiplication or exponentiation into addition or multiplication, respectively, via log properties.
• Then, differentiate implicitly with respect to the variable: Often simplifies differentiation due to split terms.

Recognizing when to use logarithmic differentiation empowers you to tackle a wide array of calculus problems much more effectively.