Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems } 1-44, \text { find } D_{x} y \text { using the rules of this section. } $$ $$ y=\frac{\alpha}{x^{3}} $$

Step-by-Step Solution

Answer

The derivative is \( D_x y = -\frac{3\alpha}{x^4} \).

Step 1: Rewrite the Function

The function given is \( y = \frac{\alpha}{x^3} \). To use the power rule for differentiation, rewrite this function as \( y = \alpha x^{-3} \).

Step 2: Apply the Power Rule

To differentiate \( y = \alpha x^{-3} \) with respect to \( x \), apply the power rule for differentiation. The power rule states that \( \frac{d}{dx}(x^n) = n \cdot x^{n-1} \).

Step 3: Differentiate the Function

Using the power rule, differentiate \( y = \alpha x^{-3} \). The derivative is \[ D_x y = \alpha \cdot (-3) x^{-3-1} \].

Step 4: Simplify the Derivative

Simplify the expression from Step 3: \( D_x y = -3\alpha x^{-4} \).

Step 5: Write the Final Answer

The derivative of \( y = \frac{\alpha}{x^3} \) with respect to \( x \) is \( D_x y = -\frac{3\alpha}{x^4} \).

Key Concepts

Power Rule

The power rule is a fundamental principle used in calculus for finding the derivative of a function. It's incredibly useful for polynomial functions and makes the differentiation process straightforward.When you have a function in the form of \( f(x) = x^n \), where \( n \) is any real number, the power rule states:

• \( \frac{d}{dx}(x^n) = n \cdot x^{n-1} \)

This means that you multiply the function by the power and then subtract one from the power. For example, if you were to differentiate \( f(x) = x^3 \), you'd get:

• \( \frac{d}{dx}(x^3) = 3 \cdot x^{2} \)

The power rule is efficient and reduces complex differentiation problems into simple steps, just as shown in the exercise when rewriting \( y = \frac{\alpha}{x^3} \) as \( y = \alpha x^{-3} \). This form allows you to apply the power rule directly.

Derivative

A derivative represents the rate at which a function is changing at any given point. It is a vital concept in calculus used to determine the slope of a function or how a function changes.In practice, when you find a derivative, you're essentially asking: how does the function \( y \) change as \( x \) changes?

• The notation \( D_x y \) refers to the derivative of \( y \) with respect to \( x \).

Using the exercise, where \( y = \alpha x^{-3} \), to find the derivative we applied the power rule and obtained \( D_x y = -3\alpha x^{-4} \). This expression tells us the behavior of the function's rate of change. A negative derivative here indicates that as \( x \) increases, \( y \) decreases.

Calculus

Calculus is a branch of mathematics that studies continuous change. It is divided into two major parts: differentiation and integration.Differentiation, as covered in your exercise, focuses on finding the derivative of functions. It provides a powerful toolkit for analyzing the behavior of functions:

• The process is primarily concerned with determining how a function's output changes with its inputs, essentially capturing the function's 'slope' at any point.
• In practical terms, differentiation can help model real-world phenomena, such as physical motion or economic trends.

By differentiating \( y = \frac{\alpha}{x^3} \), you're applying one of the two central operations in calculus. The other, integration, is its counterpart and works for finding areas under curves but is not directly addressed in your problem.