Problem 105 - Calculus and its applications

Question

Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$ y=\frac{x^{5}-x^{3}}{x^{2}} $$

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dx} \) is \( 3x^{2} - 1 \).

Step 1: Simplify the Function

Given the function \( y = \frac{x^{5} - x^{3}}{x^{2}} \), we simplify it by dividing each term in the numerator by \( x^{2} \). This gives us the expression:\[ y = x^{5-2} - x^{3-2} = x^{3} - x \]

Step 2: Differentiate the Simplified Function

Now, we differentiate the simplified function \( y = x^{3} - x \) with respect to \( x \). This requires applying the power rule, which states that \( \frac{d}{dx}(x^{n}) = nx^{n-1} \).For \( y = x^{3} - x \), the derivative is:\[ \frac{dy}{dx} = 3x^{3-1} - 1 \cdot x^{1-1} = 3x^{2} - 1 \]

Key Concepts

Algebraic Simplification

Algebraic simplification is a powerful process that makes complex equations or expressions more manageable and easier to work with, particularly when preparing to differentiate. In the exercise at hand, simplifying the given function is a critical first step. The original function is a fraction, \( y = \frac{x^5 - x^3}{x^2} \), which may initially appear complex for differentiation. To simplify, we divide each term in the numerator by the denominator term \( x^2 \), effectively rewriting the expression to enable easier arithmetic operations.

• Divide \( x^5 \) by \( x^2 \) to get \( x^{5-2} = x^3 \).
• Similarly, divide \( x^3 \) by \( x^2 \) to end up with \( x^{3-2} = x^1 = x \).

The simplified expression becomes \( x^3 - x \), which is now free of the fraction and much easier to handle in subsequent steps such as differentiation. By simplifying algebraic expressions, we reduce potential calculation errors and make the application of calculus rules straightforward.

Power Rule

The power rule is perhaps one of the simplest yet most powerful tools in finding derivatives. It allows us to differentiate polynomial terms with ease. Essentially, it applies to any term of the form \( x^n \), where \( n \) is a constant. Here's how it works:When differentiating \( x^n \), you bring down the exponent as a multiplier, and then subtract one from the exponent itself to find the derivative. The formula can be succinctly stated as:

• \( \frac{d}{dx}(x^n) = nx^{n-1} \)

In our example, the simplified function was \( y = x^3 - x \), which consists of two polynomial terms:

• For \( x^3 \), applying the power rule gives a derivative of \( 3x^{3-1} = 3x^2 \).
• For \( x^1 = x \), the derivative is simply \( 1 \cdot x^{1-1} = 1 \cdot x^0 = 1 \) since any number to the power of zero is one.

Using the power rule eliminates complex calculations and provides a streamlined method to compute derivatives efficiently and accurately.

Derivative Calculation

Calculating derivatives is a foundational concept in calculus, representing the rate of change of a function. To calculate the derivative of the function we simplified earlier, \( y = x^3 - x \), we simply take the derivatives of each term separately using the power rule.Let's break it down:

• The derivative of \( x^3 \) is \( 3x^2 \), as determined by bringing down the exponent and reducing the power by one.
• The derivative of \( -x \) is \(-1 \), which can be understood as \( -1 \times x^0 \) or simply subtracting one from a count since \( x^1 \) diminishes to zero.

Putting it all together, the derivative \( \frac{dy}{dx} \) of the function \( y = x^3 - x \) is:\[ \frac{dy}{dx} = 3x^2 - 1 \]This step-by-step calculation provides us with the instantaneous rate of change for the function. Effectively understanding and applying derivative calculation is crucial in both academic settings and real-world problem-solving scenarios, as it helps describe how systems evolve and react over time.