Question

First find and simplify $$\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ Then find \(d y / d x\) by taking the limit of your answer as \(\Delta x \rightarrow 0 .\) $$ y=\frac{1}{x+1} $$

Short answer

The derivative \(\frac{dy}{dx}\) is \(\frac{-1}{(x+1)^2}\).

Step-by-step solution

Understand the Problem

We are given a function \( y = \frac{1}{x+1} \) and need to find the difference quotient \( \frac{\Delta y}{\Delta x} = \frac{f(x+\Delta x)-f(x)}{\Delta x} \), then take the limit as \( \Delta x \to 0 \) to get the derivative \( \frac{dy}{dx} \).

Calculate \( f(x+\Delta x) \)

Substitute \( x+\Delta x \) into the function: \( f(x+\Delta x) = \frac{1}{(x+\Delta x) + 1} = \frac{1}{x + \Delta x + 1} \).

Set Up the Difference Quotient

The difference quotient \( \frac{\Delta y}{\Delta x} \) is calculated as: \( \frac{f(x+\Delta x)-f(x)}{\Delta x} = \frac{\frac{1}{x + \Delta x + 1} - \frac{1}{x + 1}}{\Delta x} \).

Simplify the Expression

Simplify the numerator: \( \frac{1}{x + \Delta x + 1} - \frac{1}{x + 1} = \frac{(x + 1) - (x + \Delta x + 1)}{(x + \Delta x + 1)(x + 1)} = \frac{-\Delta x}{(x + \Delta x + 1)(x + 1)} \).

Divide by \( \Delta x \)

After simplifying the numerator, the difference quotient becomes: \( \frac{-\Delta x}{\Delta x \cdot (x + \Delta x + 1)(x + 1)} = \frac{-1}{(x + \Delta x + 1)(x + 1)} \).

Take the Limit as \( \Delta x \to 0 \)

Taking the limit of \( \frac{-1}{(x + \Delta x + 1)(x + 1)} \) as \( \Delta x \to 0 \) gives \( \frac{dy}{dx} = \frac{-1}{(x+1)(x+1)} = \frac{-1}{(x+1)^2} \).

Key concepts

Difference Quotient

The difference quotient is a fundamental tool in calculus used to approximate the slope of the tangent line to a curve. It's like taking a small "snapshot" of a function between two points. To understand it better, envision the function as a path on a hill. The difference quotient tells us how steep the hill is between two points on this path.
The formula for the difference quotient is \[\frac{f(x+\Delta x)-f(x)}{\Delta x}\]where:

• \(f(x)\) is the function value at point \(x\).
• \(f(x+\Delta x)\) is the function value at a slightly changed point \(x+\Delta x\).
• \(\Delta x\) represents the change or the small interval between these two points.

Hence, this formula calculates the average rate of change or slope between two points on a curve. In simpler terms, it tells us how the function \(y\) changes over \(x\). For rational functions like \(y = \frac{1}{x+1}\), it allows us to find how the function behaves as \(x\) shifts by a tiny amount. This is the first step to finding the function's derivative.

Derivative

In calculus, the derivative represents the rate at which a function is changing at any given point. Think of it as measuring how quickly something happens. It's a key concept because it provides critical insights into the behavior of functions.
To find the derivative, we use the limit of the difference quotient as \(\Delta x\) approaches zero:\[ \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} \]This formula essentially shrinks the interval \(\Delta x\) until it is just a point, giving the slope of the tangent line to the curve at a specific \(x\).

• For our example with the function \(y = \frac{1}{x+1}\), after simplification, we get:
• \( \frac{dy}{dx} = \frac{-1}{(x+1)^{2}} \)

This means the slope of the tangent line changes depending on \(x\). Where the curve is steepest, the derivative is most negative, indicating rapid change. Understanding the derivative is crucial because it helps predict how a function behaves in different situations.

Limit

The concept of a limit is crucial in calculus, as it helps us understand how values approach a certain point. When we calculate the derivative, we're essentially looking at what happens to a function as \(\Delta x\) becomes infinitesimally small.
A limit examines the behavior of a function as it gets closer to a specific value. In the context of our exercise, when we take the limit of the difference quotient as \(\Delta x\) approaches zero, we're trying to find exactly how the function behaves at a precise point.

• The limit is expressed as:
• \(\lim_{\Delta x \to 0} \frac{-1}{(x + \Delta x + 1)(x + 1)}\)

Here, as \(\Delta x\) diminishes to zero, the expression inside the limit simplifies. Thus, limits are mathematical tools that help us transition from average rates of change (difference quotient) to instantaneous rates of change (derivative), providing a foundation for studying the intricate behaviors of functions.

Rational Function

In algebra, a rational function is any function that can be expressed as the quotient of two polynomials. They are essential because they appear across calculus and many real-world applications.
The function in our exercise, \(y = \frac{1}{x+1}\), is a classic example of a rational function. Its behavior is different from that of other types of functions due to possibilities of undefined points where the denominator is zero.

• Characteristics of rational functions include:
• They can have vertical asymptotes which occur where the function is undefined, such as division by zero.
• They may have horizontal asymptotes, indicating the behavior as \(x\) approaches infinity.

In calculus, understanding rational functions is vital because it helps you identify limits and derivatives, as showcased earlier in our example. Rational functions' limits and derivatives provide insight into the curves' shapes, slopes, and points where the function might not be defined.