Question

Evaluate the following derivatives. $$\frac{d}{d x}(\ln (\ln x))$$

Short answer

Answer: The derivative of the function is \(\frac{d}{d x}(\ln(\ln(x))) = \frac{1}{x \ln(x)}\).

Step-by-step solution

Identify the Outer and Inner Functions

In the given function \(f(x) = \ln(\ln(x))\), the outer function is \(\ln(u)\) and the inner function is \(u = \ln(x)\).

Determine the Derivative of the Outer Function

Differentiating the outer function \(\ln(u)\) with respect to \(u\), we get: $$\frac{d}{d u} (\ln(u)) = \frac{1}{u}$$

Determine the Derivative of the Inner Function

Differentiating the inner function \(u = \ln(x)\) with respect to \(x\), we get: $$\frac{d}{d x} (\ln(x)) = \frac{1}{x}$$

Apply Chain Rule

To find the derivative of the given function, we need to apply the chain rule: $$\frac{d}{d x}(\ln(\ln(x))) = \frac{d}{d u}(\ln(u)) \cdot \frac{d u}{d x}$$ Now substituting the derivatives found in steps 2 and 3, we get: $$\frac{d}{d x}(\ln(\ln(x))) = \frac{1}{u} \cdot \frac{1}{x}$$

Substitute Back the Inner Function

Now substitute back the inner function \(u = \ln(x)\) to the expression we found: $$\frac{d}{d x}(\ln(\ln(x))) = \frac{1}{\ln(x)} \cdot \frac{1}{x}$$ Finally, we can rewrite the expression as: $$\frac{d}{d x}(\ln(\ln(x))) = \frac{1}{x \ln(x)}$$ Hence, the derivative of the given function is: $$\boxed{\frac{d}{d x}(\ln(\ln(x))) = \frac{1}{x \ln(x)}}$$

Key concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with composite functions. It helps us find the derivative of a function that is made up of other functions. Think of it as untangling a string of functions to uncover their individual rates of change.

In essence, the chain rule states that the derivative of a composite function, say \(f(g(x))\), is the product of the derivative of the outer function \(f\) with respect to \(g\), and the derivative of the inner function \(g\) with respect to \(x\).

The formula for the chain rule is:

• If \(y = f(u)\) and \(u = g(x)\), then \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

This rule can be imagined as peeling off layers, where each layer affects the next. In our exercise example, the outer function is \(\ln(u)\), and the inner function is \(u = \ln(x)\). By differentiating each separately, and then multiplying, we effectively apply the chain rule.

Delving into Natural Logarithms

Natural logarithms are logarithms with a base of \(e\), an irrational and transcendental number approximately equal to 2.71828. The natural logarithm, written as \(\ln x\), is a way to express exponential growth and is frequently used in calculus due to its simple derivative rules.

Understanding the properties of the natural logarithm can simplify many calculus problems:

• The derivative of \(\ln x\) is \(\frac{1}{x}\), making it straightforward to differentiate.
• \(\ln(1)\) is 0 because \(e^0 = 1\).
• \(\ln(e)\) is 1 because the base \(e\) raised to the power of 1 is \(e\).
• Natural logs convert multiplicative relationships into additive ones, e.g., \(\ln(ab) = \ln a + \ln b\).

These properties make natural logarithms an invaluable tool in simplifying complex expressions, as seen in the exercise when we calculate the derivative of \(\ln(\ln(x))\).

Unpacking Function Composition

Function composition is the process of combining two functions to form a new function. It's like putting one function inside another. When dealing with composition, you essentially "plug" one function's outputs into another function.

In the context of our original exercise, we are dealing with two functions structured as \(f(x) = \ln(\ln(x))\). The function \(\ln(x)\) acts as an inner function, while \(\ln(u)\) represents the outer function with \(u = \ln(x)\). This layered structure often calls for careful analysis using the chain rule to find derivatives efficiently.

For example, in process terms:

• First, calculate \(\ln(x)\) which simplifies into the function \(u\).
• Next, apply \(\ln()\) again to the result, this time to \(u\), creating the final composition.

Recognizing and breaking down these compositions into their individual components can drastically simplify the differentiation process. It requires practice, but gets easier with recognizing patterns such as nested logarithms or trigonometric functions.