Problem 13 - 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 \(x\). $$ h(x)=\frac{2}{x} $$

Step-by-Step Solution

Answer

The derivative of \( h(x) = \frac{2}{x} \) is \( h'(x) = \frac{-2}{x^2} \).

Step 1: Understand the Function

We are tasked with finding the derivative of the function \( h(x) = \frac{2}{x} \). This is a rational function where the numerator is a constant and the denominator is a variable, \( x \).

Step 2: Write the Difference Quotient

Using the definition of the derivative \( f'(x) = \lim _{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} \), substitute \( f(x) = \frac{2}{x} \): \[ \frac{2}{x+h} - \frac{2}{x}. \]. Express the entire difference quotient \[ \lim _{h \rightarrow 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h} \].

Step 3: Simplify the Difference

Find a common denominator for the terms \( \frac{2}{x+h} \) and \( \frac{2}{x} \): The expression becomes \( \frac{2x - 2(x+h)}{x(x+h)} \). Simplify the numerator: \( 2x - 2x - 2h = -2h \). The expression is now: \[ \frac{-2h}{x(x+h)} \].

Step 4: Substitute into the Limit

Evaluate the limit: Substitute the simplified form into the difference quotient: \[ \lim _{h \rightarrow 0} \frac{-2h}{h x(x+h)} \].

Step 5: Cancel and Take the Limit

Cancel \( h \) from the numerator and denominator: \[ \lim _{h \rightarrow 0} \frac{-2}{x(x+h)} \]. Take the limit as \( h \rightarrow 0 \): \( \frac{-2}{x(x+0)} = \frac{-2}{x^2} \). The derivative \( h'(x) \) is \( \frac{-2}{x^2} \).

Key Concepts

Difference Quotient

The difference quotient is a foundational concept in calculus that helps us understand how functions change. It's used to calculate the slope of a secant line between two points on a graph. The formula for the difference quotient is:

• \( \frac{f(x+h) - f(x)}{h} \)

This formula gives us the average rate of change of the function over the interval \( h \). It's the starting point for finding the derivative of a function. In practice, you often substitute \( f(x) \) with a specific function, as we've done with \( h(x) = \frac{2}{x} \) in our problem. Once set up, the difference quotient becomes a crucial stepping stone to understanding instantaneous rates of change via derivatives.

Rational Function

A rational function is a type of function in algebra where one polynomial is divided by another. In our case, \( h(x) = \frac{2}{x} \) is a rational function where the numerator is \( 2 \), a constant, and the denominator is \( x \), a variable. Understanding rational functions is vital in calculus as they often appear in problems dealing with limits and derivatives.
To solve problems involving rational functions, you'll often require techniques such as factoring or finding a common denominator to simplify expressions. These skills are essential in calculus when dealing with derivatives and integrals, where simplifying can make otherwise complex problems more manageable.

Limit Definition of Derivative

In calculus, the derivative of a function at a certain point measures how the function's output value changes as the input changes. The limit definition of derivative formalizes this idea mathematically. The expression:

• \( f'(x) = \lim _{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} \)

represents the instantaneous rate of change, or slope, at point \( x \). By using the limit, we ensure that we're capturing the precise behavior of a function as \( h \) approaches zero, giving us the true rate of change rather than an average over a longer interval.
In the context of our problem, substituting \( h(x) = \frac{2}{x} \) into the limit definition allows us to compute the derivative rigorously and accurately. This process typically involves algebraic manipulation to simplify and carefully evaluate the limit.

Simplifying Algebraic Expressions

Simplifying algebraic expressions is a key skill in calculus, enabling us to handle complex expressions easily. In the derivative finding process, simplifying often involves:

• Finding a common denominator, as seen when dealing with the rational terms \( \frac{2}{x+h} - \frac{2}{x} \).
• Combining like terms, which can make expressions neater, such as eliminating terms like \( 2x - 2x \).
• Canceling out terms in both the numerator and denominator, like \( h \) in the expression \( \frac{-2h}{hx(x+h)} \).

These simplification techniques not only streamline the process but also allow us to accurately compute limits and derivatives. The key is to work methodically and keep track of every step to avoid errors, especially as you work with polynomials and rational expressions.