Problem 16 - 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 \(x\). $$ F(x)=\frac{x-1}{x+1} $$

Step-by-Step Solution

Answer

The derivative is \( f^{\prime}(x) = \frac{1}{(x+1)^2} \).

Step 1: Understand the Formula to Use

The derivative of a function \( f(x) \) at a point \( x \) can be found using the limit definition \( f^{\prime}(x)=\lim _{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} \). This is the formula we'll use to find the derivative of \( F(x)=\frac{x-1}{x+1} \).

Step 2: Calculate \( F(x+h) \)

Let's find \( F(x+h) \). Substitute \( x+h \) into the function: \[ F(x+h) = \frac{(x+h)-1}{(x+h)+1} = \frac{x+h-1}{x+h+1} \].

Step 3: Set Up the Derivative Formula

Set up the formula for the derivative of \( F(x) \) using the limit definition:\[ f^{\prime}(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = \lim_{h \to 0} \frac{\frac{x+h-1}{x+h+1} - \frac{x-1}{x+1}}{h} \].

Step 4: Simplify the Difference of Fractions

Combine the two fractions into a single fraction. To do this, find a common denominator:\[ \frac{x+h-1}{x+h+1} - \frac{x-1}{x+1} = \frac{(x+h-1)(x+1) - (x-1)(x+h+1)}{(x+h+1)(x+1)} \].

Step 5: Expand and Simplify the Numerator

Expand the terms in the numerator:\[(x+h-1)(x+1) - (x-1)(x+h+1) = (x^2 + hx + x - 1x - h - 1) - (x^2 + hx + x + h - x - h - 1) \].

Step 6: Simplify Further

Simplify the numerator by combining like terms.After canceling out common terms:\[hx - h(x+1) = h \] Thus, the new expression is \( h \).

Step 7: Substitute Back into Derivative Formula

Plug this back into the limit expression:\[ f^{\prime}(x) = \lim_{h \to 0} \frac{h}{h(x+h+1)(x+1)} \].

Step 8: Simplify the Limit Expression

Cancel \( h \) from the numerator and denominator:\[ f^{\prime}(x) = \lim_{h \to 0} \frac{1}{(x+h+1)(x+1)} \].

Step 9: Evaluate the Limit

Let \( h \to 0 \):\[ f^{\prime}(x) = \frac{1}{(x+1)^2} \].

Step 10: Conclude the Derivative

Thus, the derivative of the function \( F(x)=\frac{x-1}{x+1} \) is \( f^{\prime}(x) = \frac{1}{(x+1)^2} \).

Key Concepts

Limit Definition

The limit definition of a derivative is a fundamental concept in calculus. It helps us determine the instantaneous rate of change of a function at any given point. This rate of change, represented as the derivative, can be thought of as the slope of the tangent line to the function at a particular point. This definition is formally expressed as:\[f^{\prime}(x)=\lim _{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]This formula effectively measures how much the function's output value changes as we make tiny changes to the input. Calculating a derivative using the limit definition involves several steps:

• Replace \( x \) with \( x+h \) in the function to get \( f(x+h) \).
• Subtract \( f(x) \) from \( f(x+h) \) to find the difference.
• Divide the difference by \( h \) to find the average rate of change.
• Finally, take the limit as \( h \) approaches zero to find the instantaneous rate of change.

When \( h \) approaches zero, it means we are considering the smallest possible increment to see how the function behaves at the point \( x \). This makes the derivative a precise measure of a function's behavior at specific points.

Differentiation

Differentiation is the process of finding the derivative of a function. It unveils the function's behavior in terms of its rate of change. In our exercise, we used differentiation to find the derivative of a rational function given by:\[F(x)=\frac{x-1}{x+1}\]To differentiate this function using the limit definition:

• First, we calculated \( F(x+h) \) and found \( \frac{x+h-1}{x+h+1} \).
• Then, we set up the limit definition for the derivative, \( f^{\prime}(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} \).
• Next, we simplified the expression by combining the fractions and canceling terms.
• Finally, we took the limit as \( h \to 0 \), which simplified to \( f^{\prime}(x) = \frac{1}{(x+1)^2} \).

This result tells us how the function changes at any point \( x \). Differentiation allows us to explore changes, slopes, and behaviors of functions, making it essential in mathematical analysis.

Rational Functions

Rational functions are fractions where the numerator and the denominator are both polynomials. They are a fundamental component of algebra and calculus. The given function \( F(x)=\frac{x-1}{x+1} \) is an example of a rational function, where:

• The numerator is \( x-1 \).
• The denominator is \( x+1 \).

Rational functions can exhibit interesting behavior such as asymptotes, where the function doesn't exactly have a value but instead approaches infinity or a particular line. These are caused by the roots of the polynomial in the denominator.In differentiation and calculus, working with rational functions often involves simplifying complex fractions and applying limits. Such techniques are crucial in many mathematical applications, helping solve real-world problems involving rates of change, such as velocity or growth rates. By understanding the behavior of rational functions, we can predict and model scenarios effectively.