Problem 45 - Calculus and its applications

Question

Differentiate each function. \(f(x)=\frac{3 x^{2}-5 x}{x^{2}-1}\)

Step-by-Step Solution

Answer

The derivative of the function is \( f'(x) = \frac{5x^2 - 6x + 5}{(x^2 - 1)^2} \).

Step 1: Identify the Components for Differentiation

This function is a rational function in the form \( \frac{u(x)}{v(x)} \) where \( u(x) = 3x^2 - 5x \) and \( v(x) = x^2 - 1 \). We need to apply the quotient rule for differentiation.

Step 2: Recall the Quotient Rule Formula

The quotient rule states \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). We will need \( u'(x) \), \( v'(x) \), \( u(x) \), and \( v(x) \) to apply this rule.

Step 3: Differentiate the Numerator

Differentiate \( u(x) = 3x^2 - 5x \). Using basic differentiation rules, we have \( u'(x) = 6x - 5 \).

Step 4: Differentiate the Denominator

Differentiate \( v(x) = x^2 - 1 \). Using the power rule, we find \( v'(x) = 2x \).

Step 5: Apply the Quotient Rule

Substitute \( u(x) \), \( v(x) \), \( u'(x) \), and \( v'(x) \) into the quotient rule formula: \[ f'(x) = \frac{(6x - 5)(x^2 - 1) - (3x^2 - 5x)(2x)}{(x^2 - 1)^2} \]

Step 6: Simplify the Numerator

First, expand the terms in the numerator: \((6x - 5)(x^2 - 1) = 6x^3 - 6x - 5x^2 + 5\) and \((3x^2 - 5x)(2x) = 6x^3 - 10x^2\). Thus, the numerator becomes \(6x^3 - 6x - 5x^2 + 5 - 6x^3 + 10x^2\).

Step 7: Combine and Simplify the Terms

Combine like terms in the numerator: \[ -5x^2 + 10x^2 - 6x + 5 \] simplifies to \[ 5x^2 - 6x + 5 \].

Step 8: Write the Final Derivative

The differentiated function is: \[ f'(x) = \frac{5x^2 - 6x + 5}{(x^2 - 1)^2} \].

Key Concepts

Rational Functions

A rational function is a type of function that is defined by the ratio of two polynomials. In our exercise, we're working with the function \( f(x)=\frac{3x^2-5x}{x^2-1} \). Here, the numerator (\( u(x) \)) and the denominator (\( v(x) \)) are both polynomials.
This classification is important because it informs us about the method of differentiation we need to use - the quotient rule. The role of the rational function in calculus primarily involves looking at these polynomials' behavior as they approach values that could make the denominator zero, which creates undefined points or vertical asymptotes.
By studying the derivative of rational functions, students can determine the rates of change at various points, which is crucial for sketching graphs and solving optimization problems.

Basic Differentiation Rules

The process of differentiation follows a set of well-established rules known as basic differentiation rules. These are critical tools for finding derivatives and include rules like the constant rule, sum rule, and product rule.
In this exercise, we apply these rules to the polynomial components individually before using the quotient rule. For instance, when finding the derivative of the numerator, \( 3x^2 - 5x \), we leverage these basic rules to quickly discover that the derivative \( u'(x) \) is \( 6x - 5 \).
Similarly, differentiating the polynomial in the denominator \( x^2 - 1 \) yields \( v'(x) = 2x \). Understanding these rules makes the process of differentiation more manageable and aids in efficiently working through more complex functions like our rational function here.

Power Rule

The power rule is a fundamental part of basic differentiation and is frequently used when differentiating polynomials. It states that for any term \( x^n \), the derivative is \( nx^{n-1} \).
This rule is straightforward but incredibly powerful. In our exercise, we use it to differentiate \( u(x) = 3x^2 - 5x \) and \( v(x) = x^2 - 1 \), yielding derivatives \( 6x \) and \( 2x \) respectively. The power rule allows us to handle each term individually and transform complicated expressions into something simpler.
Mastering the power rule is essential for tackling a wide array of calculus problems, especially when dealing with functions that have multiple power terms in their expressions. It streamlines the process of differentiation, saving time and reducing the likelihood of mistakes.