Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\) using the rules of this section. $$ y=\frac{5 x^{2}+2 x-6}{3 x-1} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{15x^2 - 10x + 16}{(3x - 1)^2} \).

Step 1: Identify the type of function

The given function is a rational function, which is in the form of a fraction \( \frac{u(x)}{v(x)} \), where \( u(x) = 5x^2 + 2x - 6 \) and \( v(x) = 3x - 1 \).

Step 2: Recall the Quotient Rule

For a function \( y = \frac{u(x)}{v(x)} \), the derivative is given by the quotient rule: \[ \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \].

Step 3: Differentiate the numerator

Differentiate the function \( u(x) = 5x^2 + 2x - 6 \) with respect to \( x \). The derivative \( u'(x) = 10x + 2 \).

Step 4: Differentiate the denominator

Differentiate the function \( v(x) = 3x - 1 \) with respect to \( x \). The derivative \( v'(x) = 3 \).

Step 5: Apply the Quotient Rule

Substitute \( u(x), v(x), u'(x) \), and \( v'(x) \) into the quotient rule: \[ D_x y = \frac{(3x - 1)(10x + 2) - (5x^2 + 2x - 6)(3)}{(3x - 1)^2} \].

Step 6: Simplify the numerator

Calculate:- \( (3x - 1)(10x + 2) = 30x^2 + 6x - 10x - 2 = 30x^2 - 4x - 2 \)- \( (5x^2 + 2x - 6)(3) = 15x^2 + 6x - 18 \)- Thus, \( (3x - 1)(10x + 2) - (5x^2 + 2x - 6)(3) = 30x^2 - 4x - 2 - 15x^2 - 6x + 18 = 15x^2 - 10x + 16 \).

Step 7: Write the derivative

The derivative of the original function is: \[ D_x y = \frac{15x^2 - 10x + 16}{(3x - 1)^2} \].

Key Concepts

Quotient Rule

The quotient rule is a fundamental tool in calculus used for differentiating functions that are expressed as the ratio of two other functions. When you have a function of the form \( y = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable, the quotient rule provides a formula to find the derivative of \( y \). The formula is:

• \[ \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \]
  
• This rule states that the derivative of a quotient is not just the derivative of the numerator divided by the derivative of the denominator. Instead, it involves subtracting the product of the derivative of the numerator and the denominator from the product of the numerator and the derivative of the denominator, all divided by the square of the denominator.
  
• The quotient rule is crucial whenever you're working with rational functions because it systematically handles the complexity of dividing two changing quantities.

Rational Functions

Rational functions are expressions that represent the quotient of two polynomials. They appear frequently in calculus and require special rules for differentiation. A rational function takes the form \( y = \frac{u(x)}{v(x)} \), where \( u(x) \) and \( v(x) \) are polynomials. For example, the function given in the original exercise:

• \( y = \frac{5x^2 + 2x - 6}{3x - 1} \)
  
• Here, \( u(x) = 5x^2 + 2x - 6 \) is the numerator polynomial, and \( v(x) = 3x - 1 \) is the denominator polynomial.
  
• Rational functions are very significant in calculus because they often describe real-world situations where rates and ratios are involved.

Differentiation

Differentiation is the process of finding the derivative of a function. Derivatives represent how a function changes as its input changes, capturing the rate of change or the slope of the function.

• For any function \( y = f(x) \), the derivative \( f'(x) \) gives you a precise mathematical way to determine how \( y \) reacts to tiny changes in \( x \).
  
• To apply differentiation to the rational function in the exercise, we separately differentiated the numerator \( u(x) = 5x^2 + 2x - 6 \) to get \( u'(x) = 10x + 2 \) and the denominator \( v(x) = 3x - 1 \) to get \( v'(x) = 3 \).
  
• Differentiation can be thought of as the mathematics of change, and it's indispensable in calculus for analyzing functions. The application of the quotient rule is a perfect example of using differentiation to untangle the interaction of two interdependent quantities.