Problem 70 - Calculus Early Transcendentals

Question

In each part, compute \(f^{\prime}, f^{\prime \prime}, f^{\prime \prime \prime},\) and then state the formula for \(f^{(n)} .\) (a) \(f(x)=1 / x \quad\) (b) \(f(x)=1 / x^{2}\) IHint: The expression \((-1)^{n}\) has a value of 1 if \(n\) is even and \(-1 \text { if } n \text { is odd. Use this expression in your answer. }]\)

Step-by-Step Solution

Answer

(a) \( f^{(n)}(x) = \frac{(-1)^n n!}{x^{n+1}} \); (b) \( f^{(n)}(x) = \frac{(-1)^n (n+1)!}{x^{n+2}} \)."

Step 1: Find the First Derivative for Part (a)

Given the function \(f(x) = \frac{1}{x}\), we need to compute its derivative. Using the power rule for derivatives, \(f(x)\) can be rewritten as \(f(x) = x^{-1}\). The derivative is \(f'(x) = -x^{-2} = -\frac{1}{x^2}\).

Step 2: Find the Second Derivative for Part (a)

Differentiate \(f'(x) = -x^{-2}\) to find \(f''(x)\). Using the power rule again, \(f''(x) = 2x^{-3} = \frac{2}{x^3}\).

Step 3: Find the Third Derivative for Part (a)

Differentiating \(f''(x) = 2x^{-3}\) results in \(f'''(x) = -6x^{-4} = -\frac{6}{x^4}\).

Step 4: General Formula for n-th Derivative for Part (a)

The pattern emerges as: \[f^{(n)}(x) = (-1)^n \cdot n! \cdot x^{-(n+1)} = \frac{(-1)^n n!}{x^{n+1}}\]

Step 5: Find the First Derivative for Part (b)

For \(f(x) = \frac{1}{x^2}\), rewrite it as \(f(x) = x^{-2}\). Differentiate to find \(f'(x) = -2x^{-3} = -\frac{2}{x^3}\).

Step 6: Find the Second Derivative for Part (b)

Differentiate \(f'(x) = -2x^{-3}\) to obtain \(f''(x) = 6x^{-4} = \frac{6}{x^4}\).

Step 7: Find the Third Derivative for Part (b)

Differentiating \(f''(x) = 6x^{-4}\) yields \(f'''(x) = -24x^{-5} = -\frac{24}{x^5}\).

Step 8: General Formula for n-th Derivative for Part (b)

Observe the pattern: \[f^{(n)}(x) = (-1)^n \cdot (n+1)! \cdot x^{-(n+2)} = \frac{(-1)^n (n+1)!}{x^{n+2}}\]

Key Concepts

Power Rule

The Power Rule is a fundamental theorem of calculus used to find the derivative of a function in the form of a power, such as \(f(x) = x^n\). It is particularly useful because it simplifies the derivation process into a straightforward formula. The Power Rule states that:

• If \(f(x) = x^n\), then its derivative, \(f'(x)\), will be \(nx^{n-1}\).

This means we simply multiply the function by the exponent and reduce the exponent by one. To illustrate this, consider a function \(f(x) = x^{-1}\), which simplifies to \(-1x^{-2}\) when differentiated, resulting in \(-\frac{1}{x^2}\) when expressed as a fraction. Similarly, for \(f(x) = x^{-2}\), applying the Power Rule results in \(-2x^{-3}\), equivalent to \(-\frac{2}{x^3}\). This rule is the bedrock for finding derivatives and is crucial for understanding more advanced techniques like higher order derivatives.

Higher Order Derivatives

Higher order derivatives involve taking the derivative of a derivative successively. These derivatives provide deeper insights into the behavior of functions, including concavity and points of inflection. Once you understand the basic concept of a derivative, calculating higher order derivatives becomes a repetitive application of this understanding.For instance, once we derive \(f'(x) = -x^{-2}\) from \(f(x) = x^{-1}\), we take the derivative of \(f'(x)\) to find \(f''(x)\). Applying the Power Rule to \(f'(x)\), we get \(f''(x) = 2x^{-3}\). Taking it a step further for the third derivative, \(f'''(x) = -6x^{-4}\). Notice how the form and structure repeat, following a pattern dictated by calculus rules.Each derivative order reflects a statement about the geometry of the function. While the first derivative tells us about the slope, the second relates to concavity, and the third can indicate changes in the rate of concavity, among other things.

n-th Derivative Formula

The n-th derivative formula is a generalized expression for finding derivatives of any order, especially when you observe patterns in successive derivatives. With functions like \(f(x) = x^{-1}\) and \(f(x) = x^{-2}\), deriving a general formula enables us to understand any derivative order without repetitive calculations.For \(f(x) = x^{-1}\), the n-th derivative follows the formula:

• \[f^{(n)}(x) = (-1)^n \cdot n! \cdot x^{-(n+1)} = \frac{(-1)^n n!}{x^{n+1}}\]

Similarly, for \(f(x) = x^{-2}\), the general formula for the n-th derivative is:

• \[f^{(n)}(x) = (-1)^n \cdot (n+1)! \cdot x^{-(n+2)} = \frac{(-1)^n (n+1)!}{x^{n+2}}\]

These patterns emerge as you apply the Power Rule repeatedly. The factorial \(n!\) signifies the product of all positive integers up to n, representing the acceleration of growth of values. The expression \((-1)^n\) alternates signs based on whether n is even or odd, depicting the inherent oscillation in calculus applications.