Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems } 1-44, \text { find } D_{x} y \text { using the rules of this section. } $$ $$ y=\frac{2 x^{2}-1}{3 x+5} $$

Step-by-Step Solution

Answer

The derivative \( D_x y = \frac{6x^2 + 20x + 3}{(3x + 5)^2} \).

Step 1: Identify the function for differentiation

We are given the function \( y = \frac{2x^2 - 1}{3x + 5} \). Our task is to find \( D_x y \), i.e., the derivative of \( y \) with respect to \( x \). This function is in the form of a quotient \( \frac{u}{v} \).

Step 2: Apply the Quotient Rule for differentiation

The Quotient Rule states that if \( y = \frac{u}{v} \), then \( D_x y = \frac{v \cdot u' - u \cdot v'}{v^2} \). Here, let \( u = 2x^2 - 1 \) and \( v = 3x + 5 \).

Step 3: Compute the derivatives of the numerator and denominator

Calculate \( u' \) and \( v' \). For \( u = 2x^2 - 1 \), \( u' = \frac{d}{dx}(2x^2) = 4x \). For \( v = 3x + 5 \), \( v' = \frac{d}{dx}(3x) = 3 \).

Step 4: Substitute into the Quotient Rule formula

Substitute \( u \), \( v \), \( u' \), and \( v' \) into the Quotient Rule: \[ D_x y = \frac{(3x + 5)(4x) - (2x^2 - 1)(3)}{(3x + 5)^2} \].

Step 5: Simplify the expression

Distribute and simplify the terms in the numerator: \((3x + 5)(4x) = 12x^2 + 20x \) and \((2x^2 - 1)(3) = 6x^2 - 3 \). The numerator simplifies to \((12x^2 + 20x) - (6x^2 - 3) = 6x^2 + 20x + 3 \). Thus, \[ D_x y = \frac{6x^2 + 20x + 3}{(3x + 5)^2} \].

Key Concepts

Quotient Rule

In the realm of calculus, dealing with derivatives of quotients is a common and essential task. The Quotient Rule is a shortcut that makes this process much more manageable. When we have a function expressed as the division of two differentiable functions, i.e., \( y = \frac{u}{v} \), the Quotient Rule allows us to find its derivative efficiently.In simple terms, the Quotient Rule states:

• If \( y = \frac{u}{v} \), then the derivative \( D_x y \) is:
\[ D_x y = \frac{v \cdot u' - u \cdot v'}{v^2} \]

This rule requires us to:

• Differentiate the numerator \( u \) to find \( u' \)
• Differentiate the denominator \( v \) to find \( v' \)
• Substitute these derivatives into the formula above.

One important thing to remember when applying the Quotient Rule is to carefully perform each differentiation step separately to avoid mistakes. This method simplifies the process of finding derivatives for more complex rational expressions.

Calculus Problem Solving

Calculus involves various strategies for problem-solving, especially when working with derivatives. To tackle a calculus problem, like finding the derivative of a fraction, involves several step-by-step approaches. Here’s how you can solve these problems: - **Identify the problem:** Begin by clearly understanding what is given and what is required. For instance, recognizing that a function is a quotient of two expressions. - **Choose the right rule:** Applying the correct calculus rule, such as the Quotient Rule in fraction problems, simplifies the solution. - **Stepwise differentiation:** Carefully break down the function into parts (numerator and denominator) and differentiate each component separately. Once you've found the derivatives for the numerator and the denominator, it's essential to methodically substitute these into the formula and simplify. Always keep an eye on each step, ensuring that no calculation is skipped.

Derivative Calculation

Calculating derivatives is the backbone of calculus. It unravels how functions change, which helps in understanding their nature and behavior. Let's look at how derivatives are calculated for a given function.In the given problem, we need the derivatives of two functions: the numerator \( u = 2x^2 - 1 \) and the denominator \( v = 3x + 5 \). Here's how we calculate them:- **Derivative of the Numerator (\( u \))**: - The function is \( u = 2x^2 - 1 \). - The derivative \( u' \) is found by differentiating each term: - \( u' = \frac{d}{dx}(2x^2) = 4x \) (since the derivative of \( x^2 \) is \( 2x \)).- **Derivative of the Denominator (\( v \))**: - With \( v = 3x + 5 \), the derivative \( v' \) is: - \( v' = \frac{d}{dx}(3x) = 3 \) (since the derivative of a constant is \( 0 \)).Once these derivatives are in hand, substituting them back into the Quotient Rule gives us the overall derivative function. It requires meticulous attention to detail yet becomes straightforward with practice.

Function Simplification

After finding the derivative of a function using calculus rules, the next step often involves simplification. Simplification makes the expression easier to interpret and use. For the function's derivative \( D_x y \) in the original problem, simplification was necessary after application of the Quotient Rule.Let's break down the simplification process:

• The derivative obtained was:
\[ D_x y = \frac{(3x + 5)(4x) - (2x^2 - 1)(3)}{(3x + 5)^2} \]

- **Expanding terms**: Multiply out each term in the numerator. - \((3x + 5)(4x)\) expands to \(12x^2 + 20x\). - \((2x^2 - 1)(3)\) results in \(6x^2 - 3\).- **Simplifying expression**: Combine like terms in the numerator. - \( 12x^2 + 20x - 6x^2 + 3 \) simplifies to \( 6x^2 + 20x + 3 \).Finally, we write the simplified derivative as:\[ D_x y = \frac{6x^2 + 20x + 3}{(3x + 5)^2} \]This simplification helps in understanding the derivative's behavior and makes it more manageable for further analysis.