Problem 82 - Brief Applied Calculus

Question

Let \(f\) and \(g\) be differentiable functions of \(x\). Assume that denominators are not zero. True or False: \(\frac{d}{d x}\left(\frac{f}{x}\right)=\frac{x \cdot f^{\prime}-f}{x^{2}}\)

Step-by-Step Solution

Answer

True

Step 1: Recall the Quotient Rule

The quotient rule for differentiation states: if you have a function that is the division of two differentiable functions \( u(x) \) and \( v(x) \), then the derivative is given by \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \).

Step 2: Assign Functions

For the expression \( \frac{d}{dx} \left( \frac{f}{x} \right) \), consider \( u(x) = f(x) \) and \( v(x) = x \).

Step 3: Differentiate Functions

Find the derivatives: \( f'(x) \) is the derivative of \( f(x) \), and \( v'(x) = 1 \) since the derivative of \( x \) is \( 1 \).

Step 4: Apply the Quotient Rule

Substitute \( u(x), u'(x), v(x), \) and \( v'(x) \) into the quotient rule formula: \[\frac{d}{dx} \left( \frac{f}{x} \right) = \frac{x \cdot f'(x) - f(x) \cdot 1}{x^2}\]

Step 5: Simplify the Expression

Simplify the expression obtained from the quotient rule: \[\frac{x \cdot f'(x) - f(x)}{x^2}\]This matches the expression given in the statement.

Step 6: Conclude the Truth of the Statement

Since the derived formula matches the given one, the statement is true.

Key Concepts

Quotient Rule

The quotient rule is an essential tool in calculus for differentiating functions that are divided by each other. When you have two functions, say \( u(x) \) and \( v(x) \), and you want to find the derivative of their quotient, the quotient rule is your go-to formula. It's stated as:

• \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \)

This formula might look a bit complex at first, but breaking it down makes it manageable. The numerator involves two main parts:

• First, multiply the derivative of the numerator \( u'(x) \) by the denominator \( v(x) \).
• Second, subtract the product of the numerator \( u(x) \) with the derivative of the denominator \( v'(x) \).

The expression is then divided by the square of the denominator \( v(x)^2 \), ensuring that the result is a new function that describes how the original quotient function changes.

Differentiation

Differentiation is the process of finding the derivative of a function. In simple terms, it tells us how a function changes as its input changes. Imagine driving a car - differentiation helps calculate how fast your position changes over time, which is velocity.

Mathematically, differentiation involves rules and formulas. The most basic one is the power rule: if you have \( x^n \), the derivative is \( nx^{n-1} \). Other rules include the product rule, chain rule, and of course, the quotient rule that we discussed earlier.

Differentiation is crucial in understanding rates of change and in optimization problems. It helps in finding maximum or minimum values of functions by setting the derivative to zero and solving for the variable.

Derivative of a Function

A derivative of a function essentially describes its rate of change. It's like a snapshot of how a function behaves at any given point, indicating whether it's increasing, decreasing, or staying constant. Derivatives provide a quantitative measure of a function's sensitivity to change in its inputs.

The notation for the derivative is \( f'(x) \) or \( \frac{df}{dx} \). These symbols represent the same idea: the derivative of the function \( f \) with respect to \( x \).

• The derivative takes into account not just the value of the function, but also how it changes as \( x \) varies.
• This allows one to predict behaviors and trends within a mathematical model or real-life scenario.

In the exercise you encountered, the derivative calculation involved the quotient rule, showcasing the power and utility of derivatives in calculus.