Problem 34
Question
Find \(d y /\left.d x\right|_{x=-2},\) given that \(y=(x+2) / x\).
Step-by-Step Solution
Verified Answer
The derivative evaluated at \( x = -2 \) is \(-\frac{1}{2}\).
1Step 1: Understand the Problem
We are tasked with finding the derivative of the function \( y = \frac{x+2}{x} \) with respect to \( x \), and then evaluating this derivative at \( x = -2 \).
2Step 2: Simplify the Function
First, simplify the function \( y = \frac{x+2}{x} \) by dividing each term in the numerator by the denominator: \[ y = \frac{x}{x} + \frac{2}{x} = 1 + \frac{2}{x}. \]
3Step 3: Differentiate the Function
Differentiate the simplified expression \( y = 1 + \frac{2}{x} \) with respect to \( x \). \( 1 \) is a constant, so its derivative is 0. The derivative of \( \frac{2}{x} = 2x^{-1} \) is found using the power rule: \[ \frac{d}{dx}(2x^{-1}) = 2(-1)x^{-2} = -\frac{2}{x^2}. \]
4Step 4: Evaluate the Derivative at \( x = -2 \)
Substitute \( x = -2 \) into the derivative \( -\frac{2}{x^2} \): \[ -\frac{2}{(-2)^2} = -\frac{2}{4} = -\frac{1}{2}. \]
5Step 5: Conclusion
The derivative of \( \frac{x + 2}{x} \) at \( x = -2 \) is \(-\frac{1}{2}\).
Key Concepts
Derivative EvaluationPower RuleFunction SimplificationCalculus Problem Solving
Derivative Evaluation
When we talk about evaluating derivatives, we're referring to finding the rate at which a function changes at any given point. In our exercise, the goal is to find the derivative of the function \( y = \frac{x+2}{x} \) and then specifically evaluate it at \( x = -2 \). This process involves two main steps: first, we need to determine the general expression for the derivative of the function, and then we plug in \( x = -2 \) to find its specific value at that point.
The derivative formula gives us a way to understand how changes in \( x \) affect changes in \( y \). By evaluating the function at a specific point, we get a snapshot of its rate of change there, which can be crucial for understanding the behavior of the function.
The derivative formula gives us a way to understand how changes in \( x \) affect changes in \( y \). By evaluating the function at a specific point, we get a snapshot of its rate of change there, which can be crucial for understanding the behavior of the function.
Power Rule
The power rule is a handy tool in calculus used for finding the derivative of power functions. If you have a function of the form \( x^n \), the power rule tells us that the derivative is \( nx^{n-1} \).
In our example, after simplifying the function to \( y = 1 + \frac{2}{x} \), we rewrite \( \frac{2}{x} \) as \( 2x^{-1} \) to apply the power rule. Here, the exponent \( n \) is \(-1\), thus the derivative of \( 2x^{-1} \) is computed as \( 2(-1)x^{-2} \), resulting in \(-\frac{2}{x^2} \).
This method is quick and efficient, allowing us to bypass more complex differentiation techniques and directly find derivatives for functions that can be expressed as power series.
In our example, after simplifying the function to \( y = 1 + \frac{2}{x} \), we rewrite \( \frac{2}{x} \) as \( 2x^{-1} \) to apply the power rule. Here, the exponent \( n \) is \(-1\), thus the derivative of \( 2x^{-1} \) is computed as \( 2(-1)x^{-2} \), resulting in \(-\frac{2}{x^2} \).
This method is quick and efficient, allowing us to bypass more complex differentiation techniques and directly find derivatives for functions that can be expressed as power series.
Function Simplification
Before differentiating a function, simplifying it can often make the process much more straightforward.
In this exercise, we took the function \( y = \frac{x+2}{x} \) and simplified it by dividing each term in the numerator by \( x \), resulting in \( y = 1 + \frac{2}{x} \). This step helps reduce potential errors during differentiation and allows for easier manipulation of terms.
Simplifying a function first is a useful technique whether you're performing calculus operations or simply trying to better understand the structure of an equation.
In this exercise, we took the function \( y = \frac{x+2}{x} \) and simplified it by dividing each term in the numerator by \( x \), resulting in \( y = 1 + \frac{2}{x} \). This step helps reduce potential errors during differentiation and allows for easier manipulation of terms.
Simplifying a function first is a useful technique whether you're performing calculus operations or simply trying to better understand the structure of an equation.
Calculus Problem Solving
Solving calculus problems often involves multiple steps, and understanding the method to tackle each part is crucial.
In this example, we started by simplifying the function for easier differentiation, then applied the power rule to find the derivative, and finally evaluated the derivative at the specified point, \( x = -2 \). Breaking down the problem in a step-by-step manner can aid clarity and help avoid mistakes.
Successfully solving calculus problems like this one involves strategically applying mathematical rules and operations, maintaining accuracy in computations, and interpreting results to draw conclusions. By following these steps, students can cultivate a strong foundation in problem-solving within calculus.
In this example, we started by simplifying the function for easier differentiation, then applied the power rule to find the derivative, and finally evaluated the derivative at the specified point, \( x = -2 \). Breaking down the problem in a step-by-step manner can aid clarity and help avoid mistakes.
Successfully solving calculus problems like this one involves strategically applying mathematical rules and operations, maintaining accuracy in computations, and interpreting results to draw conclusions. By following these steps, students can cultivate a strong foundation in problem-solving within calculus.
Other exercises in this chapter
Problem 33
Find all values of \(x\) at which the tangent line to the given curve satisfies the stated property. \(y=\frac{x^{2}+1}{x+1} ;\) parallel to the line \(y=x\)
View solution Problem 34
Find \(d y / d x\) $$y=\frac{1+\csc \left(x^{2}\right)}{1-\cot \left(x^{2}\right)}$$
View solution Problem 34
Determine whether the statement is true or false. Explain your answer. If \(f(x)\) is a cubic polynomial, then \(f^{\prime}(x)\) is a quadratic polynomial.
View solution Problem 34
Find all values of \(x\) at which the tangent line to the given curve satisfies the stated property. \(y=\frac{x+3}{x+2} ;\) perpendicular to the line \(y=x\)
View solution