Problem 100 - University Calculus: Early Transcendentals

Question

Using mathematical induction, show that for \(n>1\) $$\frac{d^{n}}{d x^{n}} \ln x=(-1)^{n-1} \frac{(n-1) !}{x^{n}}$$

Step-by-Step Solution

Answer

The statement is true for all \(n > 1\) by induction.

Step 1: Base Case

First, we validate the base case for \(n = 2\). The given expression becomes \(\frac{d^2}{dx^2} \ln x\). Calculating the first derivative, \(\frac{d}{dx}\ln x = \frac{1}{x}\). Calculating the second derivative, \(\frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}\). This matches \((-1)^{2-1} \cdot \frac{(2-1)!}{x^2} = -\frac{1}{x^2}\). Thus, the base case holds.

Step 2: Inductive Hypothesis

Assume for \(n = k\), the formula holds: \(\frac{d^k}{dx^k} \ln x = (-1)^{k-1} \frac{(k-1)!}{x^k}\). This is our inductive hypothesis.

Step 3: Inductive Step

We need to show the statement is true for \(n = k+1\). The problem statement becomes \(\frac{d^{k+1}}{dx^{k+1}}\ln x\). Take the derivative:\[\frac{d}{dx} \left( (-1)^{k-1} \frac{(k-1)!}{x^k} \right) = (-1)^{k-1} (k-1)! \cdot \frac{-k}{x^{k+1}} = (-1)^k \frac{k!}{x^{k+1}}.\]This is the result for \(n = k+1\) as needed.

Step 4: Conclusion

As both the base case and the inductive step are verified, by mathematical induction, the statement \(\frac{d^n}{dx^n} \ln x = (-1)^{n-1} \frac{(n-1)!}{x^n}\) is true for all \(n > 1\).

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus. They represent how a function changes as its input changes. The derivative of a function at a point gives the rate at which the function value is changing at that particular point. In simpler terms, derivatives measure the slope of the function's graph at any given point.In the context of \(\ln x\), the derivative shows how the natural logarithm function grows or decreases as the value of \(x\) changes. Calculating derivatives often involves rules such as the power rule, product rule, and chain rule. These rules simplify differentiating various types of functions.

• First Derivative: Represents the initial rate of change.
• Significance: Provides insight into the behavior of functions (increasing or decreasing).

The example given shows differentiation applied multiple times, leading us into higher-order derivatives, indicative of more intricate behavior of the function.

Natural Logarithm

The natural logarithm, denoted as \(\ln x\), is the logarithm to the base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. The natural logarithm function has an array of applications, particularly prominent in math, science, and engineering.The natural log is employed in solving complex exponential equations and modeling natural growth processes like population growth or radioactive decay. In the differentiation process, the derivative of the natural logarithm function \(\ln x\) is a crucial piece, calculated as follows:

• Basic Formula: \(\frac{d}{dx} \ln x = \frac{1}{x}\)
• Characteristic: Logarithmic functions typically grow slowly, compared to exponential functions.

Understanding \(\ln x\) aids in grasping the larger picture of calculus functions, particularly when analyzing their growth patterns or converting them between different forms in the context of solving equations.

Higher-order derivatives

Higher-order derivatives extend the concept of derivatives beyond just the first. They involve taking the derivative of a derivative, repeatedly. These derivatives can offer deeper insights into the behavior of functions by analyzing acceleration, curvature, and other complex phenomena in the function's graph.For instance, the second derivative may reveal whether a function is concave up or down, which is vital in understanding points of inflection and optimizing problems.

• Notation: \(\frac{d^n}{dx^n} f(x)\)
• Application: Used in physics, engineering, and economics to study dynamics and rates of change beyond velocity, such as acceleration.
• Symmetry: Can sometimes simplify complex equations due to patterns they reveal in the derivatives.

In the given problem, higher-order derivatives of \(\ln x\) illustrate a pattern, as identified and used in the mathematical induction process. Recognizing these patterns helps establish the general formula which stands true for all integers \(n > 1\).