Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The derivative \(D_x y\) is \(\frac{4x^2 + 4x - 5}{(2x+1)^2}\).

Step 1: Identify the Function Type

The function given is a rational function, which is expressed as a fraction where the numerator is a polynomial \(2x^2 - 3x + 1\) and the denominator is another polynomial \(2x + 1\). This indicates that the Quotient Rule should be used to find the derivative.

Step 2: Recall the Quotient Rule Formula

The Quotient Rule for differentiation states that if you have a function \(y = \frac{u(x)}{v(x)}\), then its derivative \(D_x y\) is given by:\[D_x y = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}\]where \(u(x) = 2x^2 - 3x + 1\) and \(v(x) = 2x + 1\) in this case.

Step 3: Differentiate the Numerator and Denominator

Differentiate \(u(x)\) and \(v(x)\):- \(u'(x) = \frac{d}{dx}(2x^2 - 3x + 1) = 4x - 3\)- \(v'(x) = \frac{d}{dx}(2x + 1) = 2\)

Step 4: Apply the Quotient Rule

Use the Quotient Rule to find \(D_x y\):\[D_x y = \frac{(2x+1)(4x-3) - (2x^2 - 3x + 1)(2)}{(2x+1)^2}\]First, expand the numerator:\[(2x+1)(4x-3) = 8x^2 - 6x + 4x - 3 = 8x^2 - 2x - 3\]\[(2x^2 - 3x + 1)(2) = 4x^2 - 6x + 2\]Substitute and simplify to find the simplified numerator:\[8x^2 - 2x - 3 - (4x^2 - 6x + 2) = 4x^2 + 4x - 5\]

Step 5: Simplify Expression

Now we can write:\[D_x y = \frac{4x^2 + 4x - 5}{(2x+1)^2}\]

Key Concepts

Understanding Rational Functions

Rational functions are pivotal in calculus, serving as expressions where a polynomial is divided by another polynomial. In simple terms, it is a ratio of two polynomials, much like a fraction in arithmetic. The general form of a rational function is \( \frac{u(x)}{v(x)} \), where both the numerator \(u(x)\) and the denominator \(v(x)\) are polynomials.

Rational functions can behave uniquely compared to regular polynomial functions due to the division by a polynomial. One key feature is that they can have discontinuities or undefined points where the denominator is zero.

For calculus students, identifying a rational function is the first step in determining the appropriate method for differentiation or integration. In this problem, recognizing that the given function \( y = \frac{2x^2 - 3x + 1}{2x + 1} \) is rational sets the stage for applying the Quotient Rule.

Exploring Polynomial Differentiation

Polynomial differentiation involves finding the derivative of a polynomial function. This means determining how the function's output changes as its input changes. This process leverages rules like the Power Rule, which states that the derivative of \(x^n\) is \(nx^{n-1}\).

When we perform polynomial differentiation, each term is differentiated individually. For instance, in the function \(u(x) = 2x^2 - 3x + 1\), the derivatives of each term are:

• \(2x^2\) becomes \(4x\),
• \(-3x\) becomes \(-3\),
• and the constant term \(+1\) becomes \(0\).

This results in the derivative \(u'(x) = 4x - 3\).

In calculus, understanding how to differentiate polynomials is crucial, as it is foundational for more complex differentiation tasks, such as applying the Quotient Rule in rational functions.

The Derivative Calculation Process

To calculate the derivative of a rational function, we use the Quotient Rule. The rule states that if \(y = \frac{u(x)}{v(x)} \), then the derivative \( D_x y \) is:
\[ D_x y = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \]

In our problem,

• \( u(x) = 2x^2 - 3x + 1 \)
• and \( v(x) = 2x + 1 \).

We first differentiate these to obtain \(u'(x) = 4x - 3\) and \(v'(x) = 2\).

Next, we substitute the values into the Quotient Rule formula, and carry out the operations:

• First part: \((2x+1)(4x-3) = 8x^2 - 2x - 3\)
• Second part: \((2x^2 - 3x + 1)(2) = 4x^2 - 6x + 2\)

Subtracting these components results in the expression \(4x^2 + 4x - 5\), providing us with the derivative: \( \frac{4x^2 + 4x - 5}{(2x+1)^2} \).

This process showcases how calculus equips us with rules and formulas to tackle complex function differentiation.

Approach to Calculus Problem-Solving

Calculus problem-solving, especially when dealing with differentials, often requires methodical steps and a deep understanding of differentiation rules. The solution process involves a strategic approach, which starts with identifying the type of function you're dealing with.

Once identified, the next step is to recall the appropriate differentiation rule, such as the Quotient Rule in this scenario. It's crucial to separate and differentiate each part of the function accurately and then systematically apply the rule to obtain results.

Problem-solving in calculus also requires skills such as careful simplification and a focus on accuracy, especially during substitution and arithmetic operations. In approaching calculus problems, like the given rational function, students should:

• Analyze the function structure
• Choose the correct differentiation approach
• Break down the problem and perform calculations methodically
• Simplify results carefully

With practice, these problem-solving steps become intuitive, making calculus a more accessible and less daunting subject.