Derivative Of Y = √x / (1 + √x) A Calculus Problem Solved

by ADMIN 58 views

Calculus is a fundamental branch of mathematics that deals with continuous change. Differentiation, a key concept in calculus, allows us to find the rate at which a function's output changes with respect to its input. In simpler terms, it helps us determine the slope of a curve at any given point. This article delves into finding the derivative of the function y = √x / (1 + √x), a common problem in introductory calculus courses. Understanding the steps involved in solving this problem not only reinforces the application of differentiation rules but also highlights the importance of algebraic manipulation in simplifying complex expressions. The derivative, denoted as dy/dx, represents the instantaneous rate of change of y with respect to x. In practical terms, it can be used to model various phenomena, from the velocity of an object to the growth rate of a population. Therefore, mastering the techniques of differentiation is crucial for anyone pursuing studies or careers in science, engineering, economics, and related fields. Before diving into the solution, let's briefly review some essential differentiation rules that we will be using. The power rule, which states that the derivative of x^n is n * x^(n-1)*, is a cornerstone of differentiation. The quotient rule, which is particularly relevant for our problem, provides a method for differentiating functions that are expressed as a ratio of two other functions. The quotient rule states that if y = u/v, where u and v are functions of x, then dy/dx = (v(du/dx) - u(dv/dx)) / v^2. Additionally, we'll need to remember the chain rule, which is used to differentiate composite functions. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). These rules, when applied correctly, allow us to tackle a wide range of differentiation problems. Now, let's proceed with finding the derivative of y = √x / (1 + √x), step by step.

Step-by-Step Solution: Finding dy/dx

To find the derivative dy/dx for the function y = √x / (1 + √x), we will employ the quotient rule. The quotient rule is a fundamental concept in calculus, specifically in the realm of differential calculus. It provides a method for finding the derivative of a function that is expressed as the ratio of two other functions. In many mathematical and scientific contexts, functions are often represented as quotients, making the quotient rule an indispensable tool for analyzing their rates of change. Before we dive into the specifics of applying the quotient rule to our problem, let's take a moment to formally state the rule. If we have a function y that is defined as the quotient 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 dy/dx, can be calculated using the following formula: dy/dx = [v(x) * (du/dx) - u(x) * (dv/dx)] / [v(x)]^2. This formula might seem a bit daunting at first, but it essentially breaks down the process of differentiation into manageable steps. We need to identify the functions u(x) and v(x), find their respective derivatives du/dx and dv/dx, and then plug these components into the formula. The denominator of the formula involves squaring the function v(x), which is a crucial part of the quotient rule. It's important to remember this formula accurately to avoid errors in the differentiation process. The quotient rule is not just a mathematical trick; it has a solid theoretical foundation rooted in the limit definition of the derivative. It can be derived using the first principles of calculus, which involves evaluating the limit of the difference quotient as the change in x approaches zero. However, for practical applications, it's more efficient to memorize and apply the formula directly. In the context of our problem, y = √x / (1 + √x), we can clearly see that the function is expressed as a quotient. The numerator, √x, can be considered as our u(x), and the denominator, (1 + √x), can be considered as our v(x). Now that we have identified u(x) and v(x), the next step is to find their respective derivatives. This will involve applying other differentiation rules, such as the power rule and the constant rule, which we will discuss in the following steps. Once we have du/dx and dv/dx, we can plug them into the quotient rule formula and simplify the expression to find dy/dx. So, let's move on to finding the derivatives of u(x) and v(x). In our case, we identify u = √x and v = 1 + √x. This identification is the first crucial step in applying the quotient rule correctly. By recognizing the numerator and denominator as separate functions, we can proceed with finding their individual derivatives. It's important to be precise in this step, as any error in identifying u and v will propagate through the rest of the solution. The function u = √x is a simple power function, which can be rewritten as u = x^(1/2). This form is more suitable for applying the power rule of differentiation. The function v = 1 + √x is a sum of a constant and a power function. We can differentiate each term separately using the constant rule and the power rule. The constant rule states that the derivative of a constant is zero, while the power rule states that the derivative of x^n is n * x^(n-1)*. By carefully identifying u and v, we set the stage for a smooth application of the quotient rule. This step highlights the importance of understanding the structure of the given function and breaking it down into simpler components. In more complex problems, the identification of u and v might not be as straightforward, requiring a deeper understanding of function composition and algebraic manipulation. However, in this case, the functions are relatively simple, making the application of the quotient rule a clear and direct process. Now that we have identified u and v, we can move on to the next step, which is to find their respective derivatives. This will involve applying the power rule and the constant rule, as mentioned earlier. Once we have the derivatives of u and v, we can plug them into the quotient rule formula and simplify the expression to obtain the derivative of the original function, y. So, let's proceed with finding du/dx and dv/dx in the next step.

Differentiating u and v

Now, let's differentiate u and v separately. We have u = √x = x^(1/2). To find du/dx, we apply the power rule. The power rule is a fundamental differentiation rule that states if we have a function of the form f(x) = x^n, where n is a constant, then the derivative of f(x) with respect to x is given by f'(x) = n * x^(n-1). This rule is widely used in calculus and forms the basis for differentiating polynomial and power functions. The power rule is not just a formula to be memorized; it has a solid mathematical foundation derived from the limit definition of the derivative. It can be proven using the binomial theorem and the concept of limits. However, for practical applications, it's more efficient to apply the rule directly. Understanding the power rule is crucial for mastering differentiation. It allows us to find the derivatives of a wide range of functions, from simple monomials to more complex expressions involving powers and roots. The power rule is also closely related to other differentiation rules, such as the product rule and the quotient rule, which we will be using later in this problem. In the context of our problem, we have u = x^(1/2). This is a power function where the exponent n is equal to 1/2. Applying the power rule, we can find the derivative of u with respect to x. We multiply the exponent (1/2) by the function and then subtract 1 from the exponent. This gives us du/dx = (1/2) * x^((1/2) - 1). Simplifying the exponent, we get du/dx = (1/2) * x^(-1/2). The negative exponent indicates that we have a reciprocal power. We can rewrite x^(-1/2) as 1 / x^(1/2), which is equivalent to 1 / √x. Therefore, the derivative of u with respect to x is du/dx = (1/2) * (1 / √x), which can be further simplified to du/dx = 1 / (2√x). This result is a common derivative that is worth remembering. The derivative of the square root function is frequently encountered in calculus problems, so it's beneficial to have it memorized. Now that we have found du/dx, let's move on to differentiating v. We have v = 1 + √x. This function is a sum of a constant term (1) and a power function (√x). To differentiate v, we will use the constant rule and the power rule. The constant rule states that the derivative of a constant is always zero. This is because a constant function does not change its value as x changes, so its rate of change is zero. The power rule, as we discussed earlier, is used to differentiate the power function √x. By applying these rules, we can find the derivative of v with respect to x. So, let's proceed with differentiating v in the next step. For v = 1 + √x, we know that the derivative of a constant is zero, and the derivative of √x is 1/(2√x) (as we found earlier). Thus, dv/dx = 0 + 1/(2√x) = 1/(2√x). Now that we have both du/dx and dv/dx, we can substitute these into the quotient rule formula. The next step involves carefully plugging these derivatives into the quotient rule formula and simplifying the resulting expression. This step requires attention to detail and accurate algebraic manipulation. By correctly substituting the derivatives and simplifying the expression, we can arrive at the final answer for dy/dx. So, let's move on to the next step and apply the quotient rule.

Applying the Quotient Rule

Substituting into the quotient rule formula, dy/dx = [v(du/dx) - u(dv/dx)] / v^2, we get:

dy/dx = [(1 + √x)(1/(2√x)) - (√x)(1/(2√x))] / (1 + √x)^2

This step is a direct application of the quotient rule, where we plug in the expressions we found for u, v, du/dx, and dv/dx. The quotient rule, as we discussed earlier, is a fundamental differentiation rule that allows us to find the derivative of a function that is expressed as the ratio of two other functions. The formula for the quotient rule is dy/dx = [v(du/dx) - u(dv/dx)] / v^2, where u and v are functions of x. In this step, we are essentially taking the pieces we have calculated in the previous steps and assembling them according to the quotient rule formula. It's crucial to be meticulous in this substitution process to avoid errors. Each term must be placed in its correct position within the formula. The term (1 + √x), which represents v, is multiplied by du/dx, which is 1/(2√x). The term √x, which represents u, is multiplied by dv/dx, which is also 1/(2√x). The entire expression is then divided by (1 + √x)^2, which represents v^2. This substitution step is not just a mechanical process; it requires a deep understanding of the quotient rule and its components. By correctly substituting the expressions, we set the stage for simplifying the expression and arriving at the final answer. The next step involves algebraic manipulation to simplify the expression we have obtained. This will involve distributing terms, combining like terms, and possibly factoring. The simplification process is just as important as the substitution process. A complex expression can be simplified into a more manageable form, making it easier to interpret and use in further calculations. So, let's move on to the next step and simplify the expression we have obtained by applying the quotient rule. Now, we simplify the numerator:

dy/dx = [(1/(2√x) + √x/(2√x)) - √x/(2√x)] / (1 + √x)^2

This step involves simplifying the numerator of the expression we obtained after applying the quotient rule. The numerator is a combination of terms involving fractions and square roots. To simplify it, we need to perform algebraic manipulations such as distributing terms and combining like terms. The first term in the numerator is (1 + √x)(1/(2√x)). This term can be expanded by distributing the 1/(2√x) across the terms inside the parentheses. This gives us 1/(2√x) + √x/(2√x). The second term in the numerator is -(√x)(1/(2√x)), which simplifies to -√x/(2√x). Now, we have the numerator as 1/(2√x) + √x/(2√x) - √x/(2√x). We can see that the terms √x/(2√x) and -√x/(2√x) are additive inverses, meaning they cancel each other out. This leaves us with just 1/(2√x) in the numerator. This simplification step is crucial because it reduces the complexity of the expression and makes it easier to see the final result. By carefully performing the algebraic manipulations, we have transformed the numerator into a much simpler form. The simplified numerator is now 1/(2√x). The denominator, (1 + √x)^2, remains unchanged at this point. We will deal with the denominator in the next step, if necessary. Now that we have simplified the numerator, we can rewrite the entire expression as dy/dx = [1/(2√x)] / (1 + √x)^2. This expression is much cleaner and easier to work with than the original expression we obtained after applying the quotient rule. The next step is to simplify this expression further by dealing with the complex fraction. A complex fraction is a fraction where the numerator, the denominator, or both contain fractions. To simplify a complex fraction, we can multiply the numerator and the denominator by the reciprocal of the fraction in the denominator. In this case, we can rewrite the expression as a single fraction by multiplying the numerator by the reciprocal of the denominator. So, let's proceed with simplifying the complex fraction in the next step. Continuing the simplification, we have:

dy/dx = 1 / [2√x(1 + √x)^2]

This is the final simplified form of the derivative. This step is the culmination of all the previous steps, where we have meticulously applied the quotient rule, simplified the numerator, and now we are simplifying the entire expression to its most concise form. We started with a complex expression obtained after substituting the derivatives into the quotient rule formula. Through a series of algebraic manipulations, we have gradually reduced the complexity of the expression. In the previous step, we simplified the numerator and obtained 1/(2√x). We also had the denominator as (1 + √x)^2. Now, we need to combine these two parts into a single fraction. To do this, we can rewrite the expression as a division of two fractions: dy/dx = [1/(2√x)] / (1 + √x)^2. To divide by a fraction, we multiply by its reciprocal. So, we can rewrite this as dy/dx = [1/(2√x)] * [1 / (1 + √x)^2]. Now, we simply multiply the numerators and the denominators together. This gives us dy/dx = 1 / [2√x * (1 + √x)^2]. This is the final simplified form of the derivative. We have successfully found the derivative of the function y = √x / (1 + √x) using the quotient rule and algebraic simplification. The final answer, dy/dx = 1 / [2√x(1 + √x)^2], represents the instantaneous rate of change of y with respect to x. This result can be used in various applications, such as finding the slope of the tangent line to the curve at a given point, or analyzing the behavior of the function. The process of finding this derivative highlights the importance of understanding and applying the fundamental rules of calculus, as well as the ability to manipulate algebraic expressions. By mastering these skills, we can tackle a wide range of differentiation problems and apply them to solve real-world problems. In summary, we have taken the original function, applied the quotient rule, simplified the expression step by step, and arrived at the final derivative. This process demonstrates the power of calculus in analyzing functions and their rates of change. Therefore, the correct answer is:

C. 1 / [2√x(1 + √x)^2]

Conclusion

In this article, we successfully found the derivative of the function y = √x / (1 + √x) using the quotient rule and careful algebraic manipulation. This exercise demonstrates the power of calculus in determining the rate of change of functions, a crucial concept in many scientific and engineering applications. The key steps involved identifying the numerator and denominator as separate functions, applying the quotient rule formula, and simplifying the resulting expression. Each step required a solid understanding of differentiation rules and algebraic techniques. The final result, dy/dx = 1 / [2√x(1 + √x)^2], provides valuable information about the behavior of the function y with respect to changes in x. This derivative can be used to find the slope of the tangent line at any point on the curve, determine intervals of increasing and decreasing behavior, and identify local maxima and minima. The process of differentiation is not just a mechanical procedure; it requires a deep understanding of the underlying concepts and the ability to apply them creatively. By mastering these skills, students and professionals can tackle a wide range of problems in calculus and related fields. The quotient rule, in particular, is a versatile tool that can be applied to many functions expressed as ratios. It's essential to practice applying this rule in various contexts to develop fluency and accuracy. Algebraic manipulation is also a critical skill in calculus. Simplifying expressions after applying differentiation rules often requires techniques such as factoring, combining like terms, and rationalizing denominators. These skills are honed through practice and a strong foundation in algebra. In conclusion, finding the derivative of y = √x / (1 + √x) is a valuable exercise that reinforces the application of the quotient rule and algebraic manipulation. The final result provides insights into the behavior of the function and demonstrates the power of calculus in analyzing rates of change. By mastering these concepts, individuals can unlock a wide range of applications in science, engineering, and beyond.