Chapter 3: Problem 16
Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). $$ F(x)=\frac{x-1}{x+1} $$
Short Answer
Expert verified
The derivative is \( f^{\prime}(x) = \frac{1}{(x+1)^2} \).
Step by step solution
01
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 \( F(x)=\frac{x-1}{x+1} \).
02
Calculate \( F(x+h) \)
Let's find \( F(x+h) \). Substitute \( x+h \) into the function: \[ F(x+h) = \frac{(x+h)-1}{(x+h)+1} = \frac{x+h-1}{x+h+1} \].
03
Set Up the Derivative Formula
Set up the formula for the derivative of \( F(x) \) using the limit definition:\[ f^{\prime}(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = \lim_{h \to 0} \frac{\frac{x+h-1}{x+h+1} - \frac{x-1}{x+1}}{h} \].
04
Simplify the Difference of Fractions
Combine the two fractions into a single fraction. To do this, find a common denominator:\[ \frac{x+h-1}{x+h+1} - \frac{x-1}{x+1} = \frac{(x+h-1)(x+1) - (x-1)(x+h+1)}{(x+h+1)(x+1)} \].
05
Expand and Simplify the Numerator
Expand the terms in the numerator:\[(x+h-1)(x+1) - (x-1)(x+h+1) = (x^2 + hx + x - 1x - h - 1) - (x^2 + hx + x + h - x - h - 1) \].
06
Simplify Further
Simplify the numerator by combining like terms.After canceling out common terms:\[hx - h(x+1) = h \] Thus, the new expression is \( h \).
07
Substitute Back into Derivative Formula
Plug this back into the limit expression:\[ f^{\prime}(x) = \lim_{h \to 0} \frac{h}{h(x+h+1)(x+1)} \].
08
Simplify the Limit Expression
Cancel \( h \) from the numerator and denominator:\[ f^{\prime}(x) = \lim_{h \to 0} \frac{1}{(x+h+1)(x+1)} \].
09
Evaluate the Limit
Let \( h \to 0 \):\[ f^{\prime}(x) = \frac{1}{(x+1)^2} \].
10
Conclude the Derivative
Thus, the derivative of the function \( F(x)=\frac{x-1}{x+1} \) is \( f^{\prime}(x) = \frac{1}{(x+1)^2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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 \( f^{\prime}(x) = \frac{1}{(x+1)^2} \).
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 \( x+1 \).