Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). $$ G(x)=\frac{2 x-1}{x-4} $$

Step-by-Step Solution

Answer

The derivative of the function is \( G'(x) = \frac{-2}{(x - 4)^2} \).

Step 1: Define the function and its components

We are given the function \( G(x) = \frac{2x - 1}{x - 4} \). We need to find the derivative of this function using the limit definition of the derivative.

Step 2: Write the difference quotient

The difference quotient is the expression \( \frac{G(x+h) - G(x)}{h} \). First, we need to calculate \( G(x+h) \).Substitute \( x+h \) into \( G(x) \):\[ G(x+h) = \frac{2(x+h) - 1}{(x+h) - 4} = \frac{2x + 2h - 1}{x + h - 4} \] The difference quotient becomes:\[ \frac{\frac{2x + 2h - 1}{x + h - 4} - \frac{2x - 1}{x - 4}}{h} \]

Step 3: Combine the fractions

To combine the fractions, find a common denominator:The common denominator is \((x+h-4)(x-4)\). Rewrite each term in the difference quotient:\[ = \frac{(2x + 2h - 1)(x-4) - (2x-1)(x+h-4)}{h(x + h - 4)(x - 4)} \]

Step 4: Simplify the numerator

Expand and simplify the numerator:1. Expand: \[ (2x + 2h - 1)(x-4) = (2x + 2h - 1)x - (2x + 2h - 1)4 \] \[ = 2x^2 + 2hx - x - 8x - 8h + 4 \] \[ = 2x^2 + (2h-9)x - 8h + 4 \]2. Expand: \[ (2x-1)(x+h-4) = (2x-1)x + (2x-1)h - (2x-1)4 \] \[ = 2x^2 + 2xh - 4x - x - h + 4 \] \[ = 2x^2 + (2h-5)x - h + 4 \]3. Combine and simplify:Subtract the second expansion from the first:\[ 2hx + 5x - 8h + h = 8h - h = 7h \]\[ = -4x - 4 = -4x + 4 \]\[ = \frac{2hx - 8h}{(x + h - 4)(x -4)} \]

Step 5: Simplify further by canceling terms with 'h'

The resulting expression becomes:\[ = \frac{-2h}{(x + h - 4)(x -4)} \]Now, cancel \(h\) in the numerator and denominator:\[ \frac{-2}{(x + h - 4)(x -4)} \]

Step 6: Evaluate the limit as h approaches 0

Set \( h \) to approach 0 in the expression:The expression becomes: \[ \lim_{h \to 0} \frac{-2}{(x + h - 4)(x - 4)} = \frac{-2}{(x - 4)^2} \]This is the derivative \(G'(x)\).

Key Concepts

Understanding the Limit Definition of Derivative

The limit definition of a derivative is a foundational concept in calculus. Essentially, it allows us to understand how a function behaves as its input changes by an infinitesimal amount. This is depicted by the formula: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Here, \(f'(x)\) is the derivative of the function \(f(x)\). It represents the slope of the tangent line at a specific point \(x\) on the curve. The limit part, \(\lim_{h \to 0}\), tells us we are considering an infinitely small change near \(x\), and the difference quotient \(\frac{f(x+h) - f(x)}{h}\) gives us the average rate of change over interval \(h\). In our problem, we use this definition on a rational function to find how it changes at any point \(x\). Calculating the limit as \(h\) goes to zero gives us a precise rate of change, helping to derive the function's slope at that point.

Mastering the Quotient Rule

While the limit definition is a general approach, the quotient rule is a specific technique for differentiating functions expressed as one function divided by another. It comes handy for rational functions, like \(G(x) = \frac{2x - 1}{x - 4}\), where one polynomial is divided by another.The quotient rule states: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]where \(u\) and \(v\) are functions of \(x\), \(u'\) is the derivative of \(u\), and \(v'\) is the derivative of \(v\). You apply this by:

• Taking the derivative of the numerator \(u\) while keeping the denominator \(v\) intact.


• Then keep the numerator intact while taking the derivative of the denominator \(v\).


• Subtract the two results and divide by the square of \(v\).

The quotient rule helps streamline differentiation for division expressions, which could otherwise be cumbersome. In our exercise, although we used the limit definition, the quotient rule is another powerful method to understand and differentiate rational functions.

Effective Calculus Problem Solving

Calculus problem solving, especially involving limits and derivatives, requires patience and attention to detail. To successfully solve calculus problems:

• Start by clearly identifying the function and what you need to find, as shown in the exercise.


• Use systematic algebraic manipulations to simplify expressions. Combining like terms, finding common denominators, and cancelling terms are typical steps.


• Don't forget the importance of limits. Evaluating the expression as \(h\) approaches zero is crucial in finding precise derivative values.


• Always verify your steps by checking calculations for errors.

By following these strategies, you enhance your problem-solving skills, become adept at handling complex calculus questions, and further your understanding of derivatives and their applications. Practice and persistence yield mastery over time.

Differentiation of Rational Functions

Rational functions, which are ratios of two polynomials, frequently appear in calculus problems. Differentiating them often involves advanced techniques like the limit definition or quotient rule. For example, \(G(x) = \frac{2x - 1}{x - 4}\) showcases a rational function.To differentiate such functions:

• Begin by understanding the structure of the function: recognize the numerator and the denominator.


• Apply techniques such as the quotient rule or limit definition to find its derivative.


• Simplify the resulting expressions carefully to ensure clarity and correctness.

This differentiation provides insights into the behavior of rational functions, revealing aspects like slope, concavity, and overall shape of the graph. Mastery of these techniques allows deeper intuitive and practical understanding of how complex functions work.