Problem 77 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, given an example to show why it is false. If \(f\) is differentiable, then $$ \frac{d}{d x}\left[\frac{f(x)}{x^{2}}\right]=\frac{f^{\prime}(x)}{2 x} $$

Step-by-Step Solution

Answer

The statement is false. The correct derivative of \(\frac{f(x)}{x^2}\) using the Quotient Rule yields \(\frac{x^2f^{\prime}(x)-2xf(x)}{x^4}\), not \(\frac{f^{\prime}(x)}{2x}\). A counter-example can be provided using \(f(x)=x\) which results in the derivative, \(-\frac{1}{x^2}\), not equal to \(\frac{1}{2x}\).

Step 1: Understand the Problem

We have a statement that claims that the derivative of \(\frac{f(x)}{x^2}\) is equal to \(\frac{f^{\prime}(x)}{2x}\). We seek to verify if this claim is True or False.

Step 2: Apply the Derivative Rules

Taking the derivative of a quotient requires us to apply the Quotient Rule of derivations. The Quotient Rule states that the derivative of a quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all over the square of the denominator. Applying the Quotient Rule, the derivative of \(\frac{f(x)}{x^2}\), denoted by \(\frac{d}{dx}\left(\frac{f(x)}{x^2}\right)\), is: \[ \frac{x^2f^{\prime}(x)-2xf(x)}{(x^2)^2} \] We can simplify the above expression to: \[ \frac{x^2f^{\prime}(x)-2xf(x)}{x^4} \] Hence, the derivative of \(\frac{f(x)}{x^2}\) is not equal to \(\frac{f^{\prime}(x)}{2x}\). Therefore, the original statement is false.

Step 3: Provide Counter-Example

As part of the exercise, it's asked to provide an example if the statement is false. Let's consider a straightforward case where \(f(x)=x\). Thus, \[ \frac{d}{d x}\left[\frac{x}{x^{2}}\right] \] should be equal to \[ \frac{1}{2x} \] But, if we simplify the expression inside the derivative on the left-hand side, we get: \[ \frac{d}{d x}[x^{-1}]=-x^{-2}=-\frac{1}{x^2} \] And this doesn't equal \(\frac{1}{2x}\).

Key Concepts

Differentiation

Differentiation is one of the fundamental concepts in calculus, essential for understanding how functions change. When we talk about the derivative of a function, we refer to the rate at which the function's value is changing at any given point.
The derivative tells us the slope of a function at a particular point, representing the rate of change of the function value with respect to a change in its input variable. It is a powerful tool because it helps us understand and quantify the behavior of functions, like growth rates, slopes of curves, and motion.

• To find a derivative, we use the limit definition: \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).
• Differentiation can be applied in various contexts—whether in geometry for finding tangent lines or in physics for analyzing velocity and acceleration.
• Different functions have specific rules that ease the differentiation process, which brings us to the quotient rule—a prerequisite to understanding today's problem.

Quotient Rule

The Quotient Rule is an essential component of differentiation, particularly when dealing with the division of two functions. It provides a method to differentiate expressions where one function is divided by another.
If you have a function expressed as a quotient, \(\frac{u(x)}{v(x)}\), the derivative \(\frac{d}{dx}\left( \frac{u(x)}{v(x)} \right)\) can be found using this formula:

• \( \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \)

It is crucial to remember how to apply this formula step by step:

• First, differentiate the numerator \(u(x)\), giving \(u'(x)\).
• Then, differentiate the denominator \(v(x)\), obtaining \(v'(x)\).
• Multiply the denominator by the derivative of the numerator and subtract the product of the numerator and the derivative of the denominator.

Finally, divide the result by the square of the original denominator.
This rule was utilized in solving the problem to illustrate that the suggested equivalent derivative \(\frac{f'(x)}{2x}\) was incorrect as demonstrated in the step-by-step solution.

Derivative Rules

Derivative rules are sets of formulas and techniques that simplify the task of differentiation. Familiarity with these rules is crucial for efficiently solving calculus problems like the one in our exercise.
Here are a few basic derivative rules that are commonly applied:

• **Power Rule**: If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).
• **Constant Rule**: The derivative of a constant is zero, \(f'(x) = 0\) if \(f(x)\) is constant.
• **Sum Rule**: If \(f(x) = g(x) + h(x)\), then \(f'(x) = g'(x) + h'(x)\).

Advanced rules also exist for more complex operations, such as:

• **Product Rule**: Used for functions that are multiplied together, similar in principle to the quotient rule but applied to multiplication rather than division.
• **Chain Rule**: Used for composite functions—when we have a function within another function, critical for nested and complex expressions.

These rules enable us to handle a wide variety of functions and make calculations quicker while solving problems involving derivative calculations. Understanding when and how to apply each rule is key to succeeding in calculus.