Question

Find \(D_{x} y\) using the rules of this section. $$ y=\frac{2 x^{2}-1}{3 x+5} $$

Short answer

\( D_{x} y = \frac{6x^2 + 20x + 3}{(3x + 5)^2} \)

Step-by-step solution

Identify the Structure

First, recognize that the function \( y = \frac{2x^2 - 1}{3x + 5} \) is a rational function and can be differentiated using the quotient rule. The quotient rule is \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} \) where \( u \) is the numerator \( 2x^2 - 1 \) and \( v \) is the denominator \( 3x + 5 \).

Differentiate the Numerator

Find the derivative of the numerator \( u = 2x^2 - 1 \). The derivative \( u' \) using basic power rule is \( u' = \frac{d}{dx}(2x^2 - 1) = 4x \).

Differentiate the Denominator

Find the derivative of the denominator \( v = 3x + 5 \). The derivative \( v' \) is \( v' = \frac{d}{dx}(3x + 5) = 3 \) derived by using the power rule.

Apply the Quotient Rule

Use the quotient rule \( \frac{v \cdot u' - u \cdot v'}{v^2} \) to find \( D_{x} y \). Substitute the derivatives: \[ D_{x} y = \frac{(3x + 5)(4x) - (2x^2 - 1)(3)}{(3x + 5)^2} \].

Simplify the Expression

Expand and simplify the expression: \( (3x + 5)(4x) = 12x^2 + 20x \) and \( (2x^2 - 1)(3) = 6x^2 - 3 \). So, the expression becomes \[ 12x^2 + 20x - (6x^2 - 3) = 12x^2 + 20x - 6x^2 + 3 \].

Combine Like Terms

Combine the like terms in the numerator: \[ 12x^2 - 6x^2 = 6x^2 \] and simplify further to \[ 6x^2 + 20x + 3 \]. Place this over the common denominator \( (3x + 5)^2 \), resulting in \[ \frac{6x^2 + 20x + 3}{(3x + 5)^2} \].

Key concepts

Quotient Rule

The quotient rule is a powerful differentiation technique used when dealing with fractions where both the numerator and the denominator are functions of the same variable. In this problem, we have a rational function: \( y = \frac{2x^2 - 1}{3x + 5} \). Here, the numerator \( u \) is \( 2x^2 - 1 \), and the denominator \( v \) is \( 3x + 5 \).
When using the quotient rule, you need to find the derivatives of both \( u \) and \( v \). The rule itself is expressed as:

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

This formula tells us to take the derivative of the numerator \( u' \) and multiply it by the denominator \( v \), then subtract from it the product of the numerator \( u \) and the derivative of the denominator \( v' \). Finally, this expression is divided by the square of the original denominator \( v^2 \). This method is particularly useful for handling more complex rational functions.

Differentiation Techniques

Differentiation is a fundamental concept in calculus, allowing us to find the rate of change of a function. There are several techniques, but for rational functions, applying the quotient rule is key. Let's break down these steps to differentiate our function \( y = \frac{2x^2 - 1}{3x + 5} \).
First, identify the components of the rational function. As mentioned earlier, the numerator is \( u = 2x^2 - 1 \), and the denominator is \( v = 3x + 5 \).
Next, apply basic differentiation rules, like the power rule, to find the derivative of \( u \):

• For \( u = 2x^2 - 1 \), the derivative \( u' = 4x \).

Similarly, differentiate \( v \):

• For \( v = 3x + 5 \), the derivative \( v' = 3 \).

By systematically using these differentiation rules, you can find the derivatives needed to successfully apply the quotient rule. This technique simplifies the differentiation of complex fractions, making it a straightforward process to find the rate of change for complicated functions like the one in this exercise.

Power Rule

The power rule is one of the simplest and most commonly used differentiation rules. It's essential when differentiating polynomial terms in a function. The power rule states that if \( f(x) = ax^n \), then \( f'(x) = nax^{n-1} \). This rule helps us quickly find the derivative of any term where \( x \) is raised to a power.
In the context of our function \( y = \frac{2x^2 - 1}{3x + 5} \), the power rule is applied in differentiating both the numerator and the denominator.
For the numerator \( 2x^2 - 1 \), each term is differentiated separately:

• Applying the power rule to \( 2x^2 \), you multiply the exponent (2) by the coefficient (2), subtracting one from the exponent: \( 4x^{2-1} = 4x \).
• The constant term \(-1\) becomes 0, as the derivative of a constant is always 0.

Similarly, for the denominator \( 3x + 5 \), the power rule simplifies the derivative to \( 3 \), since the term \( 3x \) simply reduces to its constant multiplier.
Understanding and applying the power rule is crucial for managing the differentiation of polynomial components within rational functions. It paves the way towards effectively using the quotient rule and other differentiation techniques.