Derivative Of X√(2x-3) A Detailed Explanation With Steps

In the realm of calculus, derivatives play a pivotal role in understanding the rate at which a function's output changes with respect to its input. Mastering the technique of finding derivatives is crucial for students and professionals alike in various fields, including physics, engineering, economics, and computer science. This article delves into the process of finding the derivative of the function f(x) = x√(2x - 3). We will meticulously break down each step, providing a comprehensive understanding of the techniques involved. Our primary focus will be on applying the product rule and the chain rule, two fundamental concepts in differential calculus. These rules are essential for handling functions that are products or compositions of other functions. By the end of this discussion, you will gain a solid grasp of how to differentiate such functions, equipping you with a valuable skill for your mathematical toolkit. Furthermore, we will explore the importance of understanding the domain of the function and its derivative, ensuring that our results are mathematically sound and applicable within the correct context. This exploration will not only enhance your computational abilities but also deepen your understanding of the theoretical underpinnings of calculus.

Our objective is to find the derivative of the function f(x) = x√(2x - 3). This problem serves as an excellent example to illustrate the application of both the product rule and the chain rule in differentiation. The function f(x) is a product of two simpler functions: x and √(2x - 3). This immediately suggests the use of the product rule. Additionally, the square root term, √(2x - 3), is a composite function, meaning it's a function within a function. Specifically, it's the square root function applied to the linear function 2x - 3. Differentiating composite functions requires the chain rule. Therefore, we will need to apply both the product rule and the chain rule in a sequential manner to accurately find the derivative. This exercise not only tests our knowledge of these rules but also our ability to apply them in combination. The problem is structured to reinforce the understanding that complex functions often require a combination of differentiation techniques. As we proceed, we will highlight how each rule is applied and why it is necessary, ensuring a clear and methodical approach to solving the problem. Before diving into the differentiation process, it is crucial to recognize the domain of the function. The square root function is only defined for non-negative values, meaning 2x - 3 must be greater than or equal to zero. This consideration will be essential when interpreting the derivative and its validity.

Before we embark on the journey of finding the derivative of f(x) = x√(2x - 3), it is crucial to ensure that we have a firm understanding of the foundational concepts that underpin the process. Several key prerequisites are essential for successfully navigating this problem. First and foremost, a strong grasp of basic algebraic manipulation is indispensable. This includes the ability to simplify expressions, factorize, and work with exponents and radicals. The function we are dealing with involves a square root, so familiarity with radical expressions and their properties is particularly important. Secondly, a solid understanding of the product rule and the chain rule is non-negotiable. The product rule states that the derivative of a product of two functions is given by (uv)' = u'v + uv', where u and v are differentiable functions. The chain rule, on the other hand, is used to differentiate composite functions, and it states that the derivative of f(g(x)) is f'(g(x)) * g'(x). These rules are the workhorses of differential calculus and are fundamental to solving a wide array of problems. Furthermore, knowledge of basic differentiation rules for simple functions is required. This includes knowing the derivatives of polynomial functions (like x^n), constants, and basic trigonometric functions. In our specific problem, we will primarily use the power rule, which states that the derivative of x^n is nx^(n-1). Lastly, it is important to have a clear understanding of the concept of a derivative itself. A derivative represents the instantaneous rate of change of a function and is geometrically interpreted as the slope of the tangent line to the function's graph at a given point. Having these prerequisites firmly in place will not only facilitate the differentiation process but also enhance your overall understanding of calculus.

To find the derivative of f(x) = x√(2x - 3), we will follow a meticulous step-by-step approach, leveraging the product rule and the chain rule. This methodical approach will ensure clarity and accuracy in our calculations. First, we recognize that f(x) is a product of two functions, u(x) = x and v(x) = √(2x - 3). Therefore, we will apply the product rule, which states that (uv)' = u'v + uv'. Next, we need to find the derivatives of u(x) and v(x) individually. The derivative of u(x) = x is simply u'(x) = 1. Now, let's focus on finding the derivative of v(x) = √(2x - 3). This function is a composite function, so we will apply the chain rule. We can rewrite v(x) as (2x - 3)^(1/2). The chain rule tells us that if we have a function f(g(x)), its derivative is f'(g(x)) * g'(x). In our case, the outer function is f(u) = u^(1/2) and the inner function is g(x) = 2x - 3. The derivative of f(u) with respect to u is f'(u) = (1/2)u^(-1/2), and the derivative of g(x) with respect to x is g'(x) = 2. Applying the chain rule, we get v'(x) = (1/2)(2x - 3)^(-1/2) * 2, which simplifies to (2x - 3)^(-1/2) or 1/√(2x - 3). Now that we have u'(x) and v'(x), we can apply the product rule: f'(x) = u'v + uv' = 1 * √(2x - 3) + x * (1/√(2x - 3)).

To further simplify this expression, we need to combine the terms. We can do this by finding a common denominator. The common denominator is √(2x - 3). So, we rewrite the first term as √(2x - 3) * √(2x - 3) / √(2x - 3), which is (2x - 3) / √(2x - 3). Now, we can add the two terms: f'(x) = (2x - 3) / √(2x - 3) + x / √(2x - 3). Combining the numerators, we get f'(x) = (2x - 3 + x) / √(2x - 3), which simplifies to f'(x) = (3x - 3) / √(2x - 3). This is the derivative of f(x) = x√(2x - 3). We have successfully applied both the product rule and the chain rule to arrive at the solution. Each step was carefully executed to ensure the accuracy of the final result. This detailed walkthrough demonstrates the power and versatility of these calculus techniques in differentiating complex functions.

Let's delve into the detailed calculation of the derivative of f(x) = x√(2x - 3). This will provide a clear and thorough understanding of each step involved in the process. We begin by recognizing that f(x) is a product of two functions: u(x) = x and v(x) = √(2x - 3). According to the product rule, the derivative of a product of two functions is given by (uv)' = u'v + uv'. Therefore, we need to find the derivatives of u(x) and v(x) separately. The derivative of u(x) = x is straightforward. Applying the power rule, which states that the derivative of x^n is nx^(n-1), we find that u'(x) = 1. Now, let's tackle the derivative of v(x) = √(2x - 3). This is where the chain rule comes into play. We can rewrite v(x) as (2x - 3)^(1/2). The chain rule is used to differentiate composite functions, and it states that if we have a function f(g(x)), its derivative is f'(g(x)) * g'(x). In this case, we can identify the outer function as f(u) = u^(1/2) and the inner function as g(x) = 2x - 3. The derivative of the outer function f(u) = u^(1/2) with respect to u is f'(u) = (1/2)u^(-1/2). This is obtained by applying the power rule again. The derivative of the inner function g(x) = 2x - 3 with respect to x is g'(x) = 2. Now, we apply the chain rule: v'(x) = f'(g(x)) * g'(x) = (1/2)(2x - 3)^(-1/2) * 2. Simplifying this expression, we get v'(x) = (2x - 3)^(-1/2), which can also be written as v'(x) = 1/√(2x - 3). With both u'(x) and v'(x) in hand, we can now apply the product rule to find the derivative of f(x): f'(x) = u'v + uv' = 1 * √(2x - 3) + x * (1/√(2x - 3)).

To simplify this result, we need to combine the two terms into a single fraction. To do this, we find a common denominator, which is √(2x - 3). We rewrite the first term, √(2x - 3), as a fraction with the common denominator: √(2x - 3) = √(2x - 3) * √(2x - 3) / √(2x - 3) = (2x - 3) / √(2x - 3). Now, we can rewrite the derivative as: f'(x) = (2x - 3) / √(2x - 3) + x / √(2x - 3). Adding the fractions, we combine the numerators: f'(x) = (2x - 3 + x) / √(2x - 3). Finally, we simplify the numerator: f'(x) = (3x - 3) / √(2x - 3). This is the simplified form of the derivative of f(x) = x√(2x - 3). This detailed calculation provides a clear roadmap of the steps involved, from applying the product and chain rules to simplifying the resulting expression. Each step is justified by the fundamental principles of calculus, ensuring a rigorous and accurate solution.

When working with derivatives, it's not only essential to find the derivative itself but also to consider the domain of the function and its derivative. The domain of a function is the set of all possible input values (x-values) for which the function is defined. Understanding the domain is crucial for interpreting the results and ensuring that the solutions are mathematically valid. For the original function, f(x) = x√(2x - 3), the domain is restricted by the square root term. The expression inside the square root, 2x - 3, must be non-negative for the function to be real-valued. Therefore, we have the inequality 2x - 3 ≥ 0. Solving for x, we get 2x ≥ 3, which implies x ≥ 3/2. So, the domain of f(x) is [3/2, ∞). Now, let's consider the derivative we found, f'(x) = (3x - 3) / √(2x - 3). The domain of the derivative is also restricted by the square root in the denominator. The expression inside the square root, 2x - 3, must be positive since we cannot divide by zero. Therefore, we have the inequality 2x - 3 > 0. Solving for x, we get 2x > 3, which implies x > 3/2. Notice that the domain of the derivative is slightly different from the domain of the original function. The derivative is not defined at x = 3/2, while the original function is. This is because the square root is in the denominator of the derivative, making it undefined when 2x - 3 = 0. The domain of f'(x) is (3/2, ∞).

This distinction in domains is significant because it tells us where the derivative is valid. The derivative represents the instantaneous rate of change of the function, and this rate of change is only defined where the derivative itself is defined. In this case, the derivative f'(x) is defined for all x greater than 3/2, but not at x = 3/2. This means that while the original function f(x) has a value at x = 3/2, its rate of change is not defined at that point. Considering the domain is not just a mathematical formality; it's a critical step in understanding the behavior of the function and its derivative. It ensures that we interpret the results within the correct context and avoid making invalid conclusions. In practical applications, understanding the domain can help us identify limitations and boundaries within which our model or analysis is valid. Therefore, always remember to consider the domain when working with functions and their derivatives. It's a fundamental aspect of calculus that provides a deeper understanding of the mathematical landscape.

In this comprehensive exploration, we have successfully navigated the process of finding the derivative of the function f(x) = x√(2x - 3). This exercise served as a powerful illustration of the application of two fundamental rules in differential calculus: the product rule and the chain rule. We began by clearly stating the problem and highlighting the necessity of employing both the product and chain rules due to the function's structure as a product of two functions, one of which is a composite function. Before diving into the differentiation process, we emphasized the importance of understanding the prerequisites, including algebraic manipulation, the product rule, the chain rule, basic differentiation rules, and the concept of a derivative itself. These foundational concepts are the building blocks upon which calculus is built, and a solid grasp of them is essential for success.

We then proceeded with a step-by-step solution, meticulously breaking down each stage of the process. We first identified the two functions that form the product, u(x) = x and v(x) = √(2x - 3). We then found the derivatives of each function individually, applying the chain rule to v(x) due to its composite nature. Once we had u'(x) and v'(x), we applied the product rule to obtain the derivative of f(x). The resulting expression was then simplified by finding a common denominator and combining terms. To further enhance understanding, we provided a detailed calculation section, where each step was explicitly shown and justified. This ensures that the reader can follow the logic and reasoning behind each manipulation and application of the rules. Finally, we addressed the crucial aspect of domain consideration. We discussed the importance of determining the domain of both the original function and its derivative, highlighting the restrictions imposed by the square root term. We found that the domain of f(x) is [3/2, ∞), while the domain of f'(x) is (3/2, ∞). This distinction is critical for interpreting the derivative and understanding the points where the rate of change is defined.

In conclusion, this article has provided a thorough and detailed explanation of how to find the derivative of f(x) = x√(2x - 3). By emphasizing the underlying principles and providing step-by-step guidance, we have aimed to equip the reader with the knowledge and skills necessary to tackle similar problems in calculus. The application of the product rule and the chain rule, combined with a careful consideration of the domain, forms a powerful toolkit for differentiating a wide range of functions. This understanding is not only valuable for academic pursuits but also for practical applications in various fields where calculus plays a central role.

Derivative, Product Rule, Chain Rule, Differentiation, Calculus, Domain, Composite Function, Rate of Change, Algebraic Manipulation, Power Rule, Square Root, Function, Solution, Calculation

Finding the Derivative of x√(2x-3) A Step-by-Step Guide