Problem 43 - Calculus - AP Edition

Question

Use the quotient rule (Theorem 2.4 .2 ) to derive the formula for the derivative of \(f(x)=x^{-n},\) where \(n\) is a positive integer.

Step-by-Step Solution

Answer

The derivative is \(f'(x) = -n x^{-(n+1)}\).

Step 1: Identify the Quotient Rule

The Quotient Rule states that for two differentiable functions \(u(x)\) and \(v(x)\), the derivative of their quotient \(\frac{u(x)}{v(x)}\) is given by \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\).

Step 2: Express the Function as a Quotient

Rewrite \(f(x) = x^{-n}\) as a quotient: \(f(x) = \frac{1}{x^n} = \frac{u(x)}{v(x)}\), where \(u(x) = 1\) and \(v(x) = x^n\).

Step 3: Find the Derivatives of the Numerator and Denominator

Calculate the derivatives: \(u'(x) = 0\) (since \(u(x) = 1\) is a constant), and \(v'(x) = nx^{n-1}\) (using the power rule).

Step 4: Apply the Quotient Rule

Substitute these values into the quotient rule formula: \(f'(x) = \frac{0 \cdot x^n - 1 \cdot nx^{n-1}}{(x^n)^2}\). Simplify to get \(f'(x) = \frac{-nx^{n-1}}{x^{2n}}\).

Step 5: Simplify the Result

Further simplify \(f'(x) = \frac{-nx^{n-1}}{x^{2n}}\) using the law of exponents, \(f'(x) = -n x^{n-1-2n} = -n x^{-(n+1)}\).

Key Concepts

Derivative of Negative Powers

When dealing with functions that have negative powers, it's essential to understand how their derivatives behave. Negative exponents appear very often in calculus problems, especially when dealing with inverse relationships or reciprocal functions. In the exercise, we start with the function \(f(x) = x^{-n}\), where \(n\) is a positive integer. This can be rewritten as \(f(x) = \frac{1}{x^n}\). Negative powers often imply division or reciprocal forms, which is why this expression can be transformed into a quotient form.

• This transformation is crucial when simplifying the derivative of the function using the quotient rule.
• The key to deriving the derivative is mastering how operations on exponents work in algebra.
• Negative exponents follow the rule \(a^{-b} = \frac{1}{a^b}\), leading to simplifications as seen in the solution.

Understanding derivatives of negative powers is fundamental because it ties into several applications, such as physics and engineering, where rate of change and inverse relationships are modeled.

Power Rule in Differentiation

The power rule, a fundamental derivative rule in calculus, helps simplify the process of finding derivatives of power functions. It states that if \(f(x) = x^m\), the derivative \(f'(x) = mx^{m-1}\).

• This rule applies directly to any power function, including those with negative exponents like \(x^{-n}\).
• When applying the power rule to \(x^{-n}\), the derivative becomes \(-nx^{-n-1}\).
• This showcases how the exponent reduces by one, and the term is scaled by the original exponent value.

Using the power rule simplifies the process significantly and avoids the need for more complex rules like the quotient rule unless necessary. This understanding allows students to quickly differentiate powers, whether positive or negative, reducing expression handling complexity.

Exponential Functions in Calculus

Exponential functions are a crucial element in calculus, often appearing in both exponential growth and decay models. However, the focus here is on manipulating them in derivative problems. In the case of negative powers, although they are not exponential functions in the classical sense (such as \(e^x\)), they require exponential manipulation skills.

• Understanding negative exponents is very similar to handling exponential functions, as they both involve understanding changes in power and base operations.
• The differentiation process, especially through techniques like the quotient rule, often requires rewriting expressions using exponent rules.
• Exponential functions set the stage for more advanced calculus topics, understanding them aids in grasping logarithmic and transcendental functions.

In calculus, maintaining a firm grasp of exponential functions and manipulations allows students to tackle a wide variety of problems, ensuring they can navigate through complex expressions and simplify them as needed for differentiation.