Derivative Of Y=(2+x)/(3x+1) Explained Step-by-Step
In the realm of calculus, finding the derivative of a function is a fundamental operation. The derivative, often denoted as y', represents the instantaneous rate of change of a function with respect to its independent variable. This concept has far-reaching applications in various fields, including physics, engineering, economics, and computer science. One common type of function encountered in calculus is a rational function, which is a function expressed as the ratio of two polynomials. In this article, we will delve into the process of calculating the derivative of a rational function, using the quotient rule as our primary tool. We will illustrate this process with a specific example, providing a step-by-step guide to ensure clarity and understanding. Whether you're a student grappling with calculus concepts or a professional seeking a refresher, this article aims to equip you with the knowledge and skills to confidently tackle derivatives of rational functions. By the end of this exploration, you'll have a solid grasp of the quotient rule and its application, empowering you to solve a wide range of calculus problems. So, let's embark on this journey of mathematical discovery and unravel the intricacies of derivatives.
Understanding the Quotient Rule
At the heart of differentiating rational functions lies the quotient rule, a powerful tool that enables us to find the derivative of a function that is expressed as the quotient of two other functions. This rule is an essential component of differential calculus, providing a systematic approach to handling complex expressions. Before we dive into the specifics of our example, let's first lay a solid foundation by understanding the quotient rule itself. The quotient rule states that if we have a function y that is defined as the ratio of two functions, say u(x) and v(x), such that y = u(x) / v(x), then the derivative of y with respect to x, denoted as y', can be calculated using the following formula:
y' = [v(x) * u'(x) - u(x) * v'(x)] / [v(x)]^2
In this formula:
- u'(x) represents the derivative of the function u(x) with respect to x.
- v'(x) represents the derivative of the function v(x) with respect to x.
The quotient rule essentially tells us that to find the derivative of a quotient, we need to perform a specific combination of multiplications, subtractions, and divisions involving the original functions and their derivatives. The order of operations in this formula is crucial, as a slight deviation can lead to an incorrect result. The numerator of the quotient rule involves subtracting the product of the original numerator (u(x)) and the derivative of the denominator (v'(x)) from the product of the original denominator (v(x)) and the derivative of the numerator (u'(x)). This difference is then divided by the square of the original denominator ([v(x)]^2). Mastering the quotient rule requires not only memorizing the formula but also understanding the underlying concept and its application. It's a tool that unlocks the ability to differentiate a vast array of rational functions, making it an indispensable part of any calculus toolkit. With a firm grasp of the quotient rule, we can now confidently tackle the example at hand, applying this principle to find the derivative of the given rational function.
Applying the Quotient Rule to Our Example
Now, let's put our understanding of the quotient rule into practice by applying it to the specific example given: y = (2 + x) / (3x + 1). This function is a classic example of a rational function, where both the numerator and denominator are polynomials. To find the derivative, y', we will systematically follow the steps outlined by the quotient rule. First, we need to identify the functions u(x) and v(x). In this case, u(x) = 2 + x (the numerator) and v(x) = 3x + 1 (the denominator). The next step is to find the derivatives of u(x) and v(x). The derivative of u(x) = 2 + x with respect to x, denoted as u'(x), is simply 1. This is because the derivative of a constant (2) is 0, and the derivative of x is 1. Similarly, the derivative of v(x) = 3x + 1 with respect to x, denoted as v'(x), is 3. The derivative of 3x is 3, and the derivative of the constant 1 is 0. Now that we have u(x), v(x), u'(x), and v'(x), we can plug these values into the quotient rule formula:
y' = [v(x) * u'(x) - u(x) * v'(x)] / [v(x)]^2
Substituting the values, we get:
y' = [(3x + 1) * 1 - (2 + x) * 3] / (3x + 1)^2
This is the initial application of the quotient rule. However, we're not done yet. The next crucial step is to simplify this expression. Simplification not only makes the answer more elegant but also makes it easier to work with in subsequent calculations or applications. So, let's move on to the simplification process.
Simplifying the Expression
After applying the quotient rule, we arrived at the expression:
y' = [(3x + 1) * 1 - (2 + x) * 3] / (3x + 1)^2
The next crucial step is to simplify this expression to its most concise and manageable form. This involves performing the necessary algebraic operations to combine like terms and reduce the complexity of the numerator. Let's start by expanding the terms in the numerator:
y' = [3x + 1 - (6 + 3x)] / (3x + 1)^2
Next, we distribute the negative sign in the numerator:
y' = [3x + 1 - 6 - 3x] / (3x + 1)^2
Now, we can combine like terms in the numerator. Notice that the 3x terms cancel each other out:
y' = [1 - 6] / (3x + 1)^2
This simplifies to:
y' = -5 / (3x + 1)^2
This is the simplified form of the derivative. It's a much cleaner and more understandable expression than the one we had after initially applying the quotient rule. The simplification process is not just about aesthetics; it's about making the result more useful. A simplified expression is easier to interpret, easier to use in further calculations, and less prone to errors. In this case, we've successfully reduced the derivative to a simple fraction with a constant numerator and a squared polynomial in the denominator. This simplified form allows us to easily analyze the behavior of the derivative, such as its sign and magnitude, which can provide valuable insights into the original function. With the derivative now in its simplest form, we have successfully completed the task of finding y'. This entire process, from applying the quotient rule to simplifying the result, demonstrates the power and elegance of calculus in solving mathematical problems.
Conclusion
In this comprehensive guide, we've explored the process of finding the derivative of a rational function using the quotient rule. We began by understanding the fundamental concept of the quotient rule, which provides a systematic approach to differentiating functions expressed as the ratio of two other functions. We then applied this rule to a specific example, y = (2 + x) / (3x + 1), demonstrating each step in detail. This involved identifying the functions u(x) and v(x), finding their respective derivatives u'(x) and v'(x), and plugging these values into the quotient rule formula. However, the process didn't end there. We emphasized the importance of simplification, meticulously reducing the initial expression to its simplest form, y' = -5 / (3x + 1)^2. This final simplified form is not only more elegant but also more practical for further analysis and applications. Throughout this exploration, we've highlighted the crucial role of the quotient rule in calculus, showcasing its power in handling complex expressions. We've also underscored the significance of simplification, a step that often transforms a cumbersome expression into a manageable and insightful result. The ability to confidently apply the quotient rule and simplify the resulting expressions is a valuable skill in calculus and its related fields. It empowers you to solve a wide range of problems, from finding instantaneous rates of change to optimizing functions. Whether you're a student learning the ropes of calculus or a professional applying these concepts in your work, the knowledge and skills gained from this guide will undoubtedly prove beneficial. As you continue your journey in mathematics, remember that practice is key. The more you apply these techniques, the more proficient you'll become in using them. So, embrace the challenge, explore new problems, and continue to deepen your understanding of calculus. With dedication and perseverance, you'll unlock the power of derivatives and their applications in the world around us.
