Problem 8 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the outer and inner functions

In this exercise, the given function is \(\ln(\ln(x))\). The outer function is \(\ln(u)\), and the inner function is \(\ln(x)\). Let \(u = \ln(x)\).

Step 2: Find the derivative of the outer function

We need to find the derivative of \(\ln(u)\). The derivative of \(\ln(u)\) with respect to \(u\) is \(\frac{1}{u}\). In our case: \(\frac{d}{d u}(\ln(u)) = \frac{1}{u}\).

Step 3: Find the derivative of the inner function

Now, we need to find the derivative of the inner function, which is \(\ln(x)\). The derivative of \(\ln(x)\) with respect to \(x\) is \(\frac{1}{x}\). In our case: \(\frac{d}{d x}(\ln(x)) = \frac{1}{x}\).

Step 4: Apply the chain rule

According to the chain rule, the derivative of the composite function is the product of the derivatives of the outer and inner functions. In our case, we have: \(\frac{d}{d x}(\ln(\ln(x))) = \frac{d}{d u}(\ln(u))\cdot\frac{d}{d x}(u)\)

Step 5: Substitute the values of the derivatives

Now, we can substitute the values of the derivatives of the outer function and inner function into the equation: \(\frac{d}{d x}(\ln(\ln(x))) = \frac{1}{u}\cdot\frac{1}{x}\)

Step 6: Replace u with the original expression

We set \(u = \ln(x)\) in step 1. Now replace \(u\) with \(\ln(x)\) back in the previous equation: \(\frac{d}{d x}(\ln(\ln(x))) = \frac{1}{\ln(x)}\cdot\frac{1}{x}\)

Step 7: Simplify the expression

The final expression for the derivative of the given function is: \(\frac{d}{d x}(\ln(\ln(x))) = \frac{1}{x \ln(x)}\)

Key Concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This field constitutes a major part of modern math education. It has two major branches, differential calculus (concerning rates of change and slopes of curves) and integral calculus (concerning accumulation of quantities and the areas under and between curves).

Differential calculus introduces the concept of the derivative, which measures how a function changes as its input changes. Derivatives are fundamental for solving problems in physics, engineering, economics, and beyond, giving an exact way to quantify how one variable depends on another.

Natural Logarithm Derivatives

The derivative of the natural logarithm function (ln) has profound importance in calculus. The natural logarithm of a variable x is denoted as \(\ln(x)\), and its derivative with respect to x is given by \(\frac{d}{dx}(\ln(x)) = \frac{1}{x}\).

This formula is derived from the definition of \(e\), the base of the natural logarithms, as the unique number whose natural logarithm is 1. Therefore, the slope of the tangent line to the curve \(y = \ln(x)\) at any point is the reciprocal of the x-coordinate at that point.

Chain Rule

The chain rule is a fundamental theorem in calculus used for finding the derivative of composite functions. When we have two functions composed together, such as \(f(g(x))\), the chain rule allows us to differentiate this function by taking the derivative of the outer function (f) with respect to the inner function (g), and then multiplying it by the derivative of the inner function (g) with respect to x.

In mathematical terms, if \(h(x) = f(g(x))\), then \(h'(x) = f'(g(x)) \cdot g'(x)\). The chain rule enables complex derivatives to be taken quickly and accurately, which is especially useful in scenarios where functions are nested within each other.

Composite Functions

Composite functions occur when one function is applied within another function, effectively 'nesting' one function inside the other. This is denoted as \(f(g(x))\) where \(g\) is a function whose range is the domain of another function \(f\).

Understanding composite functions is crucial when applying the chain rule for derivatives. It is often helpful to think of composite functions as a process where the output of the inner function \(g(x)\) becomes the input for the outer function \(f\). In our exercise example, \(\ln(\ln(x))\) is a composite function with \(\ln(x)\) nested inside another logarithmic function.