Problem 34 - Calculus Early Transcendentals

Question

Find \(d y /\left.d x\right|_{x=-2},\) given that \(y=(x+2) / x\)

Step-by-Step Solution

Answer

The derivative at \( x = -2 \) is \( -\frac{1}{2} \).

Step 1: Given Function

We are given the function: \( y = \frac{x+2}{x} \). We need to differentiate this function to find \( \frac{dy}{dx} \).

Step 2: Rewrite the Function

First, rewrite the function in a more convenient form for differentiation. This can be expressed as: \( y = \frac{x+2}{x} = 1 + \frac{2}{x} \).

Step 3: Differentiate the Function

Differentiate the rewritten function with respect to \( x \).Use the power rule and the constant rule: \( \frac{dy}{dx} = 0 + \frac{d}{dx}(2x^{-1}) = -2x^{-2} = \frac{-2}{x^2} \).

Step 4: Evaluate the Derivative at \( x = -2 \)

Substitute \( x = -2 \) into the derivative: \( \frac{dy}{dx} = \frac{-2}{(-2)^2} = \frac{-2}{4} = -\frac{1}{2} \).

Key Concepts

Power Rule

The power rule is an essential technique in calculus used to find the derivative of functions. It is a straightforward method that works on terms where the variable is raised to a power. If you have a function of the form \( y = x^n \), then the power rule tells us that its derivative is \( \frac{dy}{dx} = nx^{n-1} \). This process involves:

• Multiplying the exponent by the coefficient of the term.
• Reducing the exponent by one.

For instance, in the function \( y = 2x^{-1} \), we apply the power rule as follows:- The exponent is \(-1\), so we multiply \(2\) by \(-1\) to get \(-2\).- Then, we decrease the exponent by 1, resulting in \(-2x^{-2}\).
This process transforms differentiation into a simple algebraic operation, making it easier to handle more complex functions.

Constant Rule

The constant rule is another fundamental concept in differentiation. It deals with terms that are not dependent on variables, i.e., constants. If you have a constant term \(c\), then the derivative of this term with respect to any variable \(x\) is zero. Mathematically, this is expressed as:

• \( \frac{d}{dx}(c) = 0 \)

Using this rule in our function, \( y = 1 + \frac{2}{x} \), let's focus on the constant term \(1\).
Since it does not include any variable, its derivative is simply zero. By knowing the constant rule, we streamline finding the derivative of mixed terms, where some components are constants.

Derivative Evaluation

Derivative evaluation refers to substituting a specific value into the derivative of a function to find the rate of change at that particular point. This step is essential once you have determined the derivative, as it provides insight into the behavior of the function at a specific input.In our exercise, after finding the derivative to be \( \frac{dy}{dx} = \frac{-2}{x^2} \), we evaluate this at \( x = -2 \). Here are the steps:

• Substitute \( x = -2 \) into the expression \( \frac{-2}{x^2} \).
• Calculate \( (-2)^2 \) which results in \(4\).
• Thus, \( \frac{-2}{4} \) simplifies to \( -\frac{1}{2} \).

This approach gives us the slope of the tangent line or the rate at which \(y\) changes with respect to \(x\) at \(x = -2\). This calculated value, \(-\frac{1}{2}\), provides concrete data about the function's instantaneous behavior at the specified point.