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 \(\Delta x \rightarrow 0 .\) $$ y=1+\frac{1}{x} $$

Short answer

\( \frac{dy}{dx} = \frac{-1}{x^2} \).

Step-by-step solution

Identify the Functions

First, recognize the given function as \[ f(x) = 1 + \frac{1}{x} \].Your task is to find \( \frac{\Delta y}{\Delta x} \) which involves finding \( f(x + \Delta x) \).

Find \( f(x + \Delta x) \)

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

Substitute into the Difference Quotient

Now substitute \( f(x + \Delta x) \) and \( f(x) \) into the difference quotient:\[ \frac{\Delta y}{\Delta x} = \frac{\left(1 + \frac{1}{x + \Delta x}\right) - \left(1 + \frac{1}{x}\right)}{\Delta x} \].

Simplify the Difference Quotient

Cancel out the constant 1 from the numerator:\[ \frac{\Delta y}{\Delta x} = \frac{\frac{1}{x + \Delta x} - \frac{1}{x}}{\Delta x} \].Find a common denominator for the fractions in the numerator:\[ \frac{1}{x + \Delta x} - \frac{1}{x} = \frac{x - (x + \Delta x)}{x(x + \Delta x)} = \frac{-\Delta x}{x(x + \Delta x)} \].Thus, the fraction becomes:\[ \frac{\Delta y}{\Delta x} = \frac{\frac{-\Delta x}{x(x + \Delta x)}}{\Delta x} \].

Simplify by Canceling Terms

Simplifying further by canceling \( \Delta x \) gives:\[ \frac{\Delta y}{\Delta x} = \frac{-1}{x(x + \Delta x)} \].

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

To find \( \frac{dy}{dx} \), take the limit as \( \Delta x \to 0 \):\[ \lim_{\Delta x \to 0} \frac{-1}{x(x + \Delta x)} = \frac{-1}{x^2} \].

Key concepts

Difference Quotient

The difference quotient is a formula used in calculus to determine the average rate of change of a function over a specific interval.
It is expressed as: \[ \frac{\Delta y}{\Delta x} = \frac{f(x+\Delta x) - f(x)}{\Delta x} \] This formula measures how much the function's output, \(\Delta y\), changes as its input, \(\Delta x\), changes.
It provides an approximation of the slope of the tangent line to the function at a given point.

The difference quotient is pivotal for understanding the concept of a derivative, as it forms the basis for its calculation.
By applying this formula to a function, we can analyze how its output changes in direct response to shifts in its input, giving us insights into the function's behavior.

Limit of a Function

Calculating the limit of a function is crucial in calculus as it gives us the value that a function approaches as its input gets infinitely close to a particular point.
In the context of the difference quotient, as \(\Delta x\) approaches zero, we're essentially zeroing in on the exact rate of change of the function.

When you reach the final stage of calculating \(\frac{dy}{dx}\), you take the limit of the simplified difference quotient: \[ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \] By doing this, you're able to find the derivative, which is the instantaneous rate of change.
This step is important as it transforms the difference quotient from an average measure to a precise measurement, helping us derive exact values from functions.

Rate of Change

The rate of change is a fundamental concept that informs us how one quantity changes in relation to another.
In calculus, this is often about how a function \(y\) changes as its input \(x\) changes, and it's typically expressed using derivatives.

Understanding the rate of change is vital in a variety of fields, such as physics for velocity and acceleration, in economics for cost and revenue changes, and in biology for growth rates.
When you compute \(\frac{dy}{dx}\), you're determining how quickly the value of \(y\) changes with respect to \(x\) at a very specific point.
This gives insights into the behavior and trends depicted by the function and helps in making predictions or understanding phenomena accurately.

Difference Quotient Simplification

Simplifying the difference quotient is an important step in effectively using it to find derivatives.
By breaking down the expression, we isolate terms and remove redundancies, making the process much easier.

This involves:

• Canceling similar terms to avoid unnecessary complexity.
• Finding common denominators to simplify fractions.
• Canceling out \(\Delta x\) when possible to refine the expression to its most straightforward form.

For example, in the original exercise, the simplification step took the difference quotient from a complex fraction to a simpler form by factoring and reducing terms.
As a result, it prepared the expression for the limit process, ultimately leading to a cleaner, more manageable derivative result.
Mastering this method is essential for anyone delving into calculus, as it enables more precise and efficient calculation of derivatives.