Problem 33 - College Algebra and Calculus: An Applied Approach

Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ f(x)=\frac{3-2 x-x^{2}}{x^{2}-1} $$

Step-by-Step Solution

Answer

The derivative of the function \( f(x)=\frac{3-2 x-x^{2}}{x^{2}-1} \) is \( f'(x) = \frac{6x^{3} - 3x^{2} - 6x + 2}{(x^{2} - 1)^{2}} \)

Step 1: Identify the functions for differentiation

First, identify the numerator and the denominator as individual functions that will be differentiated. In this case:1. The numerator, the function \( g(x) = 3 - 2x - x^{2} \)2. The denominator, the function \( h(x) = x^{2} - 1 \)

Step 2: Differentiate the individual functions

Find the derivatives of \( g \) and \( h \):1. The derivative of \( g(x) = 3 - 2x - x^{2} \) is \( g'(x) = -2 - 2x \)2. The derivative of \( h(x) = x^{2} - 1 \) is \( h'(x) = 2x \)

Step 3: Apply the Quotient Rule

Next, apply the Quotient Rule, substituting \( g, g', h, h' \) with the functions and their derivatives:According to the Quotient Rule:\[f'(x) = \frac{g'h - gh'}{h^{2}}\]Substituting, we have:\[f'(x) = \frac{(-2 - 2x) \cdot (x^{2} - 1) - (3 - 2x - x^{2}) \cdot 2x}{(x^{2} - 1)^{2}}\]

Step 4: Simplify the result

Lastly, simplify the expression to get the derivative of the function:\[f'(x) = \frac{-2x^{2} + 2 + 2x^{3} - x^{2} - 6x + 4x^{3}}{(x^{2} - 1)^{2}} = \frac{6x^{3} - 3x^{2} - 6x + 2}{(x^{2} - 1)^{2}}\]

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental tool in calculus for finding derivatives of functions presented as a fraction, or quotient, of two differentiable functions. Specifically, if you have a function in the form \( f(x) = \frac{g(x)}{h(x)} \), the Quotient Rule comes into play. To apply it, first recognize the top part, \( g(x) \), as the numerator, and the bottom part, \( h(x) \), as the denominator. It involves differentiating both the numerator and the denominator independently.According to the Quotient Rule:

• Find the derivatives: \( g'(x) \) for the numerator and \( h'(x) \) for the denominator.
• Then apply the formula: \[f'(x) = \frac{g'(x) h(x) - g(x) h'(x)}{(h(x))^2}\]

This formula helps compute the derivative of the entire quotient function \( f(x) \). The key to correct application is maintaining the order of terms in the formula, followed by thorough simplification.

Differentiation

Differentiation is the central operation in calculus that allows us to find the rate at which a function changes at any given point. When you differentiate a function, you are essentially calculating its derivative - a new function that gives this rate of change.The differentiation process relies on understanding elementary derivative rules:

• The Power Rule, which helps differentiate polynomials efficiently.
• Product, Chain, and Quotient Rules, each serving different types of functions' operations ( products, composites, and quotients, respectively).

In our exercise, the derivatives were found using these guidelines:- For the numerator \( g(x) \), which was \( 3 - 2x - x^2 \), we applied basic differentiation rules to get \( g'(x) = -2 - 2x \).- For the denominator \( h(x) = x^2 - 1 \), differentiating gives \( h'(x) = 2x \).These derivatives are the building blocks needed for the Quotient Rule.

Calculus Concepts

Calculus is a branch of mathematics that studies change. Through concepts like differentiation and integration, it helps in understanding how quantities vary with respect to one another. In this context, the concept of a derivative offers profound insights. The derivative of a function at any given point can tell us:

• The slope of the tangent line at that point.
• The instantaneous rate of change of the function.

Understanding calculus concepts means grasping not just how to compute derivatives but also their applications: - They are used extensively in sciences and engineering to model behavior, optimize processes, and predict changes. - In economics, derivatives can indicate marginal cost or revenue. The exercise in question illustrates applying these concepts practically, deepening the understanding of calculus' power in shedding light on dynamic systems.