Problem 14 - Calculus - AP Edition

Question

$$\text { Find } f^{\prime}(x)$$. $$f(x)=\frac{2 x^{2}+5}{3 x-4}$$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{6x^2 - 16x - 15}{(3x-4)^2} \).

Step 1: Identify the Function

The function given is \( f(x) = \frac{2x^2 + 5}{3x - 4} \). This is a rational function where the numerator is \( 2x^2 + 5 \) and the denominator is \( 3x - 4 \).

Step 2: Recognize the Rule

We need to find the derivative \( f'(x) \). For a function \( \frac{u}{v} \), where \( u = 2x^2 + 5 \) and \( v = 3x - 4 \), we apply the quotient rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \).

Step 3: Differentiate the Numerator

Calculate the derivative of the numerator \( u = 2x^2 + 5 \). The derivative \( u' = \frac{d}{dx}(2x^2 + 5) = 4x \).

Step 4: Differentiate the Denominator

Calculate the derivative of the denominator \( v = 3x - 4 \). The derivative \( v' = \frac{d}{dx}(3x - 4) = 3 \).

Step 5: Apply the Quotient Rule

Substitute \( u' = 4x \), \( v = 3x - 4 \), \( u = 2x^2 + 5 \), and \( v' = 3 \) into the quotient rule. This gives us: \( f'(x) = \frac{(4x)(3x - 4) - (2x^2 + 5)(3)}{(3x - 4)^2} \).

Step 6: Simplify the Expression

Expand and simplify the expression: \[ f'(x) = \frac{12x^2 - 16x - (6x^2 + 15)}{(3x - 4)^2} = \frac{12x^2 - 16x - 6x^2 - 15}{(3x - 4)^2} = \frac{6x^2 - 16x - 15}{(3x - 4)^2} \].

Step 7: Final Answer

The derivative of the function \( f(x) = \frac{2 x^{2}+5}{3 x-4} \) is \[ f'(x) = \frac{6x^2 - 16x - 15}{(3x-4)^2} \].

Key Concepts

Quotient Rule

The quotient rule is a fundamental tool in calculus for finding the derivative of a quotient of two functions. Essentially, it's used when you have a ratio or division between two differentiable functions, which are often represented as \( \frac{u}{v} \). To find the derivative \( \frac{d}{dx}\left(\frac{u}{v}\right) \), the quotient rule states it's equal to \( \frac{u'v - uv'}{v^2} \). Breaking it down:

• \( u \) is the numerator function.
• \( v \) is the denominator function.
• \( u' \) is the derivative of the numerator.
• \( v' \) is the derivative of the denominator.

The quotient rule helps solve derivatives for complex rational functions by transforming them into simpler algebraic expressions. It's crucial to ensure both \( u \) and \( v \) are differentiable and remember that the denominator \( v \) must never be zero, as division by zero is undefined.

Rational Functions

Rational functions are expressions that represent the quotient of two polynomials, such as \( \frac{2x^2 + 5}{3x - 4} \). These functions are common in calculus and involve polynomial division. They have important properties, including:

• The domain of a rational function is all real numbers except where the denominator is zero, as the function would be undefined at those points.
• The behavior at vertical asymptotes and horizontal asymptotes can provide valuable information about the graph of the function and its limits.

Understanding rational functions involves looking at their asymptotic behavior and critical points. When differentiating rational functions, we often use the quotient rule, given that their structure inherently involves division.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the function's rate of change. It is one of the core operations in calculus and is pivotal for understanding how functions behave. Key points about differentiation include:

• It involves finding the slope of the tangent line to the curve at any point.
• Common rules of differentiation include the power rule, product rule, and quotient rule.
• Derivatives provide information about the function's critical points, such as maxima, minima, and points of inflection.

For rational functions, differentiating involves using rules like the quotient rule to handle the division of polynomials. Differentiation transforms complex functions into manageable expressions that reveal the instantaneous rate of change at any point.

Calculus

Calculus is a branch of mathematics that studies continuous change. It comprises differentiation and integration, which are concerned with rates of change and areas under curves, respectively. Calculus allows us to model and solve problems in fields like physics, engineering, and economics. In the context of this exercise:

• Differentiation (one-half of calculus) helps analyze the rate of change of functions like \( f(x) = \frac{2x^2 + 5}{3x - 4} \).
• Applying calculus concepts helps understand dynamic systems and transition from simple straight-line problems to more complex curve-related challenges.
• It’s often concerned with understanding behavior at limits, such as finding derivatives at critical points or using integration to compute area under curves.

Mastering calculus requires familiarity with its various techniques, including rules like the quotient rule, and concepts like limits, which together help tackle a wide array of practical and theoretical problems.