Finding The Derivative Of Y = (2x^2 - 1)(-x^3 - X)
Introduction
In the realm of calculus, finding the derivative of a function is a fundamental operation. The derivative, denoted as dy/dx, represents the instantaneous rate of change of a function y with respect to its variable x. This concept is crucial in various fields, including physics, engineering, economics, and computer science, as it allows us to analyze the behavior of functions and model real-world phenomena. In this comprehensive guide, we will delve into the process of determining the derivative of the function y = (2x^2 - 1)(-x^3 - x). We will explore different approaches, including the product rule and the distributive property, to arrive at the correct solution. This exploration will not only provide a step-by-step solution but also enhance your understanding of derivative calculations and their applications.
Understanding the Problem
The given function, y = (2x^2 - 1)(-x^3 - x), is a product of two expressions: (2x^2 - 1) and (-x^3 - x). To find the derivative, we can employ the product rule, a fundamental concept in calculus. The product rule states that the derivative of a product of two functions is equal to the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function. Mathematically, this can be expressed as:
d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
where u(x) and v(x) are differentiable functions of x, and u'(x) and v'(x) represent their respective derivatives. Alternatively, we can first expand the product using the distributive property and then differentiate the resulting polynomial. Both methods are valid and will lead to the same answer. The choice of method often depends on personal preference or the complexity of the functions involved. In this case, we will demonstrate both methods to provide a comprehensive understanding of the derivative calculation process. Understanding the problem thoroughly is the first step towards finding the correct solution and appreciating the underlying mathematical principles.
Method 1: Applying the Product Rule
Step 1: Identify u(x) and v(x)
The first step in applying the product rule is to identify the two functions that are being multiplied. In our case, we have:
- u(x) = 2x^2 - 1
- v(x) = -x^3 - x
Step 2: Find the derivatives of u(x) and v(x)
Next, we need to find the derivatives of u(x) and v(x) with respect to x. Using the power rule, which states that d/dx(x^n) = nx^(n-1), we can find the derivatives as follows:
- u'(x) = d/dx (2x^2 - 1) = 4x
- v'(x) = d/dx (-x^3 - x) = -3x^2 - 1
Step 3: Apply the Product Rule Formula
Now, we can apply the product rule formula:
d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
Substituting the values we found in the previous steps, we get:
dy/dx = (4x)(-x^3 - x) + (2x^2 - 1)(-3x^2 - 1)
Step 4: Simplify the Expression
Finally, we need to simplify the expression by expanding the products and combining like terms:
dy/dx = -4x^4 - 4x^2 + (-6x^4 - 2x^2 + 3x^2 + 1)
dy/dx = -4x^4 - 4x^2 - 6x^4 - 2x^2 + 3x^2 + 1
dy/dx = -10x^4 - 3x^2 + 1
Therefore, the derivative of y = (2x^2 - 1)(-x^3 - x) with respect to x, using the product rule, is dy/dx = -10x^4 - 3x^2 + 1. This method provides a direct approach to finding the derivative of a product of two functions, emphasizing the importance of the product rule in calculus.
Method 2: Expanding and Differentiating
Step 1: Expand the Product
An alternative method to find the derivative is to first expand the product using the distributive property, also known as the FOIL method (First, Outer, Inner, Last). This involves multiplying each term in the first expression by each term in the second expression:
y = (2x^2 - 1)(-x^3 - x)
y = 2x2(-x3) + 2x^2(-x) - 1(-x^3) - 1(-x)
y = -2x^5 - 2x^3 + x^3 + x
Step 2: Simplify the Expression
Now, we simplify the expression by combining like terms:
y = -2x^5 - x^3 + x
Step 3: Differentiate the Polynomial
Next, we differentiate the simplified polynomial with respect to x. We apply the power rule to each term: d/dx(x^n) = nx^(n-1):
dy/dx = d/dx (-2x^5 - x^3 + x)
dy/dx = -10x^4 - 3x^2 + 1
Step 4: Compare the Results
As we can see, the result obtained by expanding and differentiating is the same as the result obtained using the product rule. This confirms the consistency of different methods in calculus. This method demonstrates that expanding the product before differentiation can sometimes simplify the process, especially when dealing with complex expressions. It also reinforces the fundamental principles of polynomial differentiation and the power rule.
Final Result
Both methods, the product rule and expanding and differentiating, lead to the same result. Therefore, the derivative of y = (2x^2 - 1)(-x^3 - x) with respect to x is:
dy/dx = -10x^4 - 3x^2 + 1
This result represents the instantaneous rate of change of the function y with respect to x. It is a polynomial function of degree 4, indicating a more complex behavior compared to the original function. Understanding the derivative allows us to analyze the function's increasing and decreasing intervals, find critical points, and determine concavity, providing a comprehensive understanding of the function's behavior.
Conclusion
In this comprehensive guide, we have successfully determined the derivative of the function y = (2x^2 - 1)(-x^3 - x) using two different methods: the product rule and expanding and differentiating. Both methods demonstrate the fundamental principles of calculus and lead to the same result: dy/dx = -10x^4 - 3x^2 + 1. The product rule is a powerful tool for differentiating products of functions, while expanding and differentiating can be advantageous for simplifying expressions before differentiation. The choice of method often depends on the specific problem and personal preference. Understanding these techniques is crucial for mastering calculus and applying it to various real-world problems. The derivative, as a fundamental concept, allows us to model and analyze rates of change, optimize functions, and gain insights into the behavior of complex systems. This exploration not only provides a solution to the given problem but also enhances the reader's understanding of calculus and its applications.
