Problem 14 - Calculus and its applications

Question

Differentiate two ways: first, by using the Quotient Rule; then, by dividing the expressions before differentiating. Compare your results as a check. Use a graphing calculator to check your results. \(y=\frac{x^{7}}{x^{3}}\)

Step-by-Step Solution

Answer

The derivative of \( y = \frac{x^7}{x^3} \) is \( 4x^3 \) by both methods.

Step 1: Identify the Function for Differentiation

We are given the function \( y = \frac{x^7}{x^3} \) and need to find its derivative using two methods.

Step 2: Use the Quotient Rule

The Quotient Rule states that if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \). Here, \( u = x^7 \) and \( v = x^3 \).Calculate \( \frac{du}{dx} = 7x^6 \) and \( \frac{dv}{dx} = 3x^2 \).Applying the Quotient Rule:\[ \frac{dy}{dx} = \frac{x^3 (7x^6) - x^7 (3x^2)}{x^6} = \frac{7x^9 - 3x^9}{x^6} = \frac{4x^9}{x^6} = 4x^3. \]

Step 3: Simplify the Function and Differentiate

Before differentiating, simplify the function:\[ y = \frac{x^7}{x^3} = x^{7-3} = x^4. \]Now, differentiate the simplified function using the Power Rule:\( \frac{dy}{dx} = 4x^{4-1} = 4x^3. \)

Step 4: Compare the Results

Both methods give the same derivative:Quotient Rule yields \( 4x^3 \) and Simplifying then Differentiating yields \( 4x^3 \).

Step 5: Verify with a Graphing Calculator

Check the derivative \( 4x^3 \) by graphing the derivative on a calculator and confirming it matches the provided function's derivative graph.

Key Concepts

Differentiation

Differentiation is the mathematical process of finding the derivative of a function. Derivatives represent the rate at which a quantity changes. It's like asking how fast something is speeding up or slowing down at a specific moment. This concept is fundamental in calculus and appears in physics, engineering, and economics.
In the given exercise, differentiation helps us find how the function \( y = \frac{x^7}{x^3} \) changes as \( x \) changes. By finding the derivative, we can identify the function's slope at any point. The exercise also emphasizes understanding concepts by using two computational methods: the Quotient Rule and simplifying first before applying the Power Rule.
Understanding differentiation ensures you can assess dynamic systems effectively, be it tracking the population growth of a species or predicting how fast stock prices might fluctuate. The goal is to make sense of change across various disciplines.

Power Rule

The Power Rule is a shortcut in calculus for finding derivatives of power functions. If you have a function where \( y = x^n \), the Power Rule tells us that the derivative is \( \frac{dy}{dx} = nx^{n-1} \). This rule significantly simplifies differentiation tasks and is especially useful when dealing with polynomial expressions.
In the original exercise, after simplifying \( y = \frac{x^7}{x^3} \) to \( y = x^4 \), we can swiftly apply the Power Rule to find the derivative. Using the Power Rule, the derivative of \( x^4 \) is \( 4x^3 \), which matches the result obtained using the Quotient Rule.
This example demonstrates the Power Rule's power—pun intended—in providing quick solutions to derivatives. It's a time-saver in calculus when expressions can be reduced to power functions, transforming complex tasks into manageable ones.

Derivative Verification

After computing derivatives using different methods, it's crucial to verify that your results are correct. Ensuring accuracy can involve comparing outcomes and using additional tools, like a graphing calculator.
In our exercise, the student employs two methods to find the derivative of the given function. Both approaches yield \( 4x^3 \), reinforcing that the solutions are consistent and accurate. To further solidify understanding, a graphing calculator visually validates the expected derivative curve, assuring that both calculation methods align with the graph.
Verification offers peace of mind and trust in your computations, demonstrating proficiency in calculus and cementing foundational skills essential in more advanced mathematical analysis. This step is particularly important in professional fields where precision and accuracy can lead to successful outcomes and solutions.