Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(y=1 / x\). Find the value of \(d y\) in each case. (a) \(x=1, d x=0.5\) (b) \(x=-2, d x=0.75\)

Step-by-Step Solution

Answer

(a) \( dy = -0.5 \); (b) \( dy = -0.1875 \).

Step 1: Write down the function and differentiate

The given function is \( y = \frac{1}{x} \). To find \( dy \), we first need to differentiate \( y \) with respect to \( x \). Therefore, we have: \[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{x} \right) = -\frac{1}{x^2}. \]

Step 2: Use the differentiation to find dy for (a)

For the case when \( x = 1 \) and \( dx = 0.5 \), we use the differential formula: \[ dy = \frac{dy}{dx} \, dx = -\frac{1}{x^2} \, dx. \]Substitute \( x = 1 \) and \( dx = 0.5 \) into the formula: \[ dy = -\frac{1}{1^2} \cdot 0.5 = -0.5. \]

Step 3: Use the differentiation to find dy for (b)

For the case when \( x = -2 \) and \( dx = 0.75 \), again use the differential formula: \[ dy = \frac{dy}{dx} \, dx = -\frac{1}{x^2} \, dx. \]Substitute \( x = -2 \) and \( dx = 0.75 \) into the formula: \[ dy = -\frac{1}{(-2)^2} \cdot 0.75 = -0.1875. \]

Key Concepts

Differential Calculus

Differential calculus is a fundamental tool in mathematics that deals with the study of how things change. Its primary goal is to understand and calculate the rate of change of functions, which is essentially the slope of the function at any point.
Understanding differential calculus involves concepts like derivatives, which provide a way to calculate the instantaneous rate of change. This is essential in fields like physics and engineering, where it is crucial to understand how one variable affects another over time.

• The derivative can be thought of as the "speed" of the function, indicating how rapidly it changes.
• It is also used to find the slope of a tangent line to the function at any given point.
• The process of finding a derivative is known as differentiation, which is central to differential calculus.

When we talk about differential calculus in the context of the given example, we use it to find how exactly the variable \( y \) in the function \( y = \frac{1}{x} \) changes with respect to \( x \). Differentiation allows us to compute these rates of change effectively.

Derivative Calculation

The calculation of derivatives is a critical aspect of mathematics that helps in understanding how a function behaves. Derivatives are calculated by applying rules of differentiation to a given function.
The derivative of a function expresses how fast the function's value changes as its input changes, and it is denoted by \( \frac{dy}{dx} \) in symbols.In the example provided, we have the function \( y = \frac{1}{x} \). To find its derivative, we use the rule of power functions, leading to:\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}.\]
This result tells us how the function \( y \) changes for each unit change in \( x \). The negative sign indicates that \( y \) decreases as \( x \) increases.

• Derivative calculations help in determining the slope or steepness of the curve represented by the function.
• They also allow us to find maximum or minimum points in functions, essential for optimization problems.

Mathematical techniques like the product, chain, and quotient rules are often used to simplify the process of derivative calculation.

Applications of Derivatives

Derivatives have a wide range of applications across various fields, making them a powerful tool in mathematics. They enable us to solve practical problems by analyzing rates of change.
In our exercise example, the derivative \( -\frac{1}{x^2} \) was used to solve problems involving small changes in \( x \). By knowing the derivative, we determine how a small increase in \( x \) (\( dx \)) affects the change in \( y \) (\( dy \)).

• In physics, derivatives describe velocity, acceleration, and other rate-related quantities.
• In economics, derivatives help in understanding and predicting changes in trends such as demand and cost.
• In engineering, they are used to model systems and understand stability conditions.

For the specific function \( y = \frac{1}{x} \), applying derivatives tells us that small positive changes in \( x \) result in typically negative changes in \( y \), emphasizing the reciprocal relationship. This is particularly valuable when precise adjustments and predictions are required.