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In Exercises 13 through \(24,\) use the quotient rule to find the derivative. $$ f(x)=\frac{2-3 x}{2 x-1} $$

Short Answer

Expert verified
The derivative of \( f(x) = \frac{2 - 3x}{2x - 1} \) is \( \frac{-1}{(2x - 1)^2} \).

Step by step solution

01

Identify the Functions

In the quotient \( \frac{u}{v} \), identify the numerator \( u(x) = 2 - 3x \) and the denominator \( v(x) = 2x - 1 \).
02

Derivatives of Functions

Find the derivatives of the numerator and denominator functions.\( u'(x) = -3 \) is the derivative of \( u(x) = 2 - 3x \), and \( v'(x) = 2 \) is the derivative of \( v(x) = 2x - 1 \).
03

Apply Quotient Rule Formula

The quotient rule is \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \). Substitute \( u'(x) = -3 \), \( v'(x) = 2 \), \( u(x) = 2 - 3x \), and \( v(x) = 2x - 1 \) into the formula.
04

Simplify the Expression

Calculate \( u'v = -3(2x - 1) = -6x + 3 \) and \( uv' = (2 - 3x)(2) = 4 - 6x \). Then substitute these into the quotient rule formula to get: \( \frac{-6x + 3 - (4 - 6x)}{(2x - 1)^2} \).
05

Further Simplification

Simplify the numerator: \( -6x + 3 - 4 + 6x = -1 \). The final derivative is \( \frac{-1}{(2x - 1)^2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivatives
Derivatives are a fundamental concept in calculus that measure how a function changes as its input changes. They provide us with valuable insight into the rate of change of a function. In simpler terms, a derivative represents the slope of the tangent line to the curve of a function at a given point.
The quotient rule is a technique specifically applied to find the derivative when you have a function that is a fraction of two other functions. Understanding the derivatives of both the numerator and denominator functions is crucial when applying the quotient rule. This is because the derivatives tell you how the top part (numerator) and the bottom part (denominator) of the fraction are changing.
For a function expressed as \( \frac{u}{v} \), where \( u \) and \( v \) are functions of \( x \), you'd need the derivatives \( u'(x) \) and \( v'(x) \) to proceed effectively using the quotient rule. As seen in our exercise, determining \( u'(x) = -3 \) and \( v'(x) = 2 \) are key steps to solving for the derivative effectively.
Numerator and Denominator Functions
In the context of differentiation using the quotient rule, identifying the numerator and denominator functions correctly sets the stage for successful application of the rule. Each component of the rational function has a specific role.
The numerator function, denoted as \( u(x) \), is the top part of the fraction. In the exercise, this is represented by \( 2 - 3x \). By finding its derivative, \( u'(x) = -3 \), we learn how this part of the function changes in response to small changes in \( x \).
The denominator function, denoted as \( v(x) \), is the bottom part of the fraction, here \( 2x - 1 \). Its derivative, \( v'(x) = 2 \), informs us of its rate of change. This information is critical as the denominator impacts how steeply the whole rational function increases or decreases.
Understanding both functions individually allows you to effectively manage their interactions through the quotient rule.
Simplification of Expressions
Simplifying expressions is the final and crucial step in differentiation, especially when using the quotient rule. After applying the quotient rule and obtaining the derivative expression, it's important to simplify it to make it more understandable and manageable.
The simplification process involves combining like terms and reducing the expression where possible. As seen in our step-by-step solution, we calculated \( u'v - uv' \) to get the expression \( -6x + 3 - (4 - 6x) \).
The simplification of these terms results in a much easier expression: \( -1 \). Finally, the entire derivative becomes \( \frac{-1}{(2x - 1)^2} \).
By simplifying, the derivative is transformed into a form that is not only easier to read but also simpler to use in further mathematical processes, such as finding slopes at specific points or solving optimization problems. Simplification reduces potential mistakes in calculations and clarifies the behavior of the function in question.

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