Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find a formula for \(D_{x}^{n}(1 / x)\).

Step-by-Step Solution

Answer

The formula is \((-1)^n \cdot n! \cdot x^{-(n+1)}\).

Step 1: Understand the Problem

We need to find the nth derivative of the function \ \( f(x) = \frac{1}{x} \). This problem requires applying the rules of differentiation repeatedly to arrive at a general formula for the nth derivative.

Step 2: Compute the First Few Derivatives

Start by calculating the first few derivatives of \( f(x) = \frac{1}{x} \). \[f'(x) = \frac{d}{dx}\left( x^{-1} \right) = -x^{-2} = -\frac{1}{x^2}\]\[f''(x) = \frac{d}{dx}\left( -x^{-2} \right) = 2x^{-3} = \frac{2}{x^3}\]\[f'''(x) = \frac{d}{dx}\left( 2x^{-3} \right) = -6x^{-4} = -\frac{6}{x^4}\]Compute more derivatives if necessary, until you see a pattern.

Step 3: Identify the Pattern

Observe the signs and coefficients from the derivatives:- The first derivative has coefficient \(-1\).- The second derivative has coefficient \(2\).- The third derivative has coefficient \(-6\).The pattern seems to involve alternating signs and factorials: \(-1\), \(2 = 2\cdot1\), \(-6 = -3\cdot2\). This is not exactly factorials, but a pattern emerges with Hamilton numbers.

Step 4: Express nth Derivative Formula

The nth derivative \( f^{(n)}(x) \) can be expressed using factorial and alternating sign patterns:\[f^{(n)}(x) = (-1)^n \cdot n! \cdot x^{-(n+1)}\]The formula involves \((-1)^n\) for alternating signs, \(n!\) for the coefficient pattern, and \(x^{-(n+1)}\) indicating the power increase by \(1\) with each derivative.

Key Concepts

Differentiation Rules

Differentiation rules are fundamental in calculus, allowing us to find the rate at which a function is changing at any point. When dealing with the function \( f(x) = \frac{1}{x} \), we need to apply differentiation rules to find its derivatives. The most common rules include:

• Power rule: For a function \( x^n \), the derivative is \( n \cdot x^{n-1} \).
• Product rule: For a product of two functions, \( f(x) \cdot g(x) \), the derivative is \( f'(x) \cdot g(x) + f(x) \cdot g'(x) \).
• Quotient rule: For a quotient of two functions, \( \frac{f(x)}{g(x)} \), the derivative is \( \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{g(x)^2} \).

In the case of \( f(x) = \frac{1}{x} \), we express it as a power function \( x^{-1} \). Using the power rule simplifies the differentiation process. As we differentiate, we notice a pattern with alternating signs and increasing negative powers. Differentiation rules help us systematically apply these formulas to find higher-order derivatives.

Factorials

Factorials are a mathematical operation defined as the product of all positive integers up to a given number \( n \). Denoted as \( n! \), factorials increase very rapidly. For example:

• \( 0! = 1 \)
• \( 1! = 1 \)
• \( 2! = 2 \times 1 = 2 \)
• \( 3! = 3 \times 2 \times 1 = 6 \)

In the problem at hand, factorials appear in the coefficients of the derivatives. As we differentiate \( \frac{1}{x} \), we observe the factorial pattern in the coefficients: -1, 2 (which can be seen as \( 2! \)), and -6 (which can be seen as \( 3! \)). This relates to the nth derivative formula: \((-1)^n \cdot n! \cdot x^{-(n+1)}\).
This formula indicates the coefficient follows the factorial operation, helping capture the complexity of changes as the function is repeatedly differentiated.

Power Functions

Power functions are expressions of the form \( x^n \), where \( n \) is any real number. They are simple yet powerful components of calculus. They allow us to easily apply differentiation rules such as the power rule. In the case of differentiating \( \frac{1}{x} \), we express it as \( x^{-1} \). Each differentiation decreases the exponent by 1, due to the power rule, leading to an expression of the form \( x^{-(n+1)} \) in the nth derivative.
For example, starting with \( f(x) = x^{-1} \):

• The first derivative is \( f'(x) = -x^{-2} \).
• The second derivative becomes \( f''(x) = 2x^{-3} \).
• By the third derivative, we have \( f'''(x) = -6x^{-4} \).

Hence, with each differentiation, the power reduces, creating a more negative exponent. This results in the alteration pattern observed in the nth derivative formula, determining the nature of the change in the function at every level of differentiation.