Implicit Differentiation Solving For Y'(9) Given Root X + Root Y = 7

Jul 10, 2025 by ADMIN 69 views

$Exploring Implicit Differentiation to Solve $\sqrt{x} + \sqrt{y} = 7$ and Find $y'(9)$$

Implicit differentiation is a powerful technique in calculus that allows us to find the derivative of a function, particularly when it's defined implicitly rather than explicitly. In this article, we'll delve into a problem where we're given the equation $\sqrt{x} + \sqrt{y} = 7$ and the condition $y(9) = 16$ , and our goal is to find $y'(9)$ using implicit differentiation. This exploration will not only demonstrate the mechanics of implicit differentiation but also highlight its practical applications in solving related rates problems and understanding the behavior of implicitly defined functions. Let's embark on this mathematical journey and unravel the intricacies of this fascinating concept.

Understanding Implicit Differentiation

Implicit differentiation is a cornerstone of calculus, particularly when dealing with equations that define functions implicitly. Unlike explicit functions, where y is directly expressed in terms of x (e.g., y = f(x)), implicit functions involve equations where x and y are intertwined, making it difficult or impossible to isolate y. Consider equations like x² + y² = 25 or, as in our problem, $\sqrt{x} + \sqrt{y} = 7$ . These equations define a relationship between x and y without explicitly giving y as a function of x.

The core idea behind implicit differentiation is to differentiate both sides of the equation with respect to x, treating y as a function of x. This necessitates the use of the chain rule, a fundamental concept in calculus that dictates how to differentiate composite functions. When we encounter a term involving y, we differentiate it with respect to y and then multiply by dy/dx, which represents the derivative of y with respect to x. This is because y is not an independent variable but rather a function of x. This process allows us to find dy/dx, which represents the rate of change of y with respect to x, even when we cannot explicitly solve for y. The power of implicit differentiation lies in its ability to handle complex relationships between variables, providing a versatile tool for solving a wide range of calculus problems. Its applications extend beyond finding derivatives to solving related rates problems, analyzing the behavior of curves, and understanding the dynamics of systems described by implicit equations. In essence, implicit differentiation unlocks a deeper understanding of functions and their derivatives, expanding the scope of calculus and its applications.

Problem Statement: $\sqrt{x} + \sqrt{y} = 7$ and $y(9) = 16$

Let's delve into the specifics of our problem. We are given the equation $\sqrt{x} + \sqrt{y} = 7$ , which implicitly defines a relationship between x and y. Our objective is to find y'(9), which represents the derivative of y with respect to x evaluated at x = 9. This means we need to determine the instantaneous rate of change of y as x changes, specifically at the point where x is equal to 9. Additionally, we are provided with the condition y(9) = 16. This tells us that when x is 9, y is 16. This information is crucial because it gives us a specific point on the curve defined by the equation $\sqrt{x} + \sqrt{y} = 7$ , which we will use to find the value of y'(9) after we have found the general expression for dy/dx.

The equation $\sqrt{x} + \sqrt{y} = 7$ can be visualized as a curve in the xy-plane. The condition y(9) = 16 tells us that the point (9, 16) lies on this curve. Finding y'(9) essentially means finding the slope of the tangent line to the curve at this point. This has significant geometric interpretation and is fundamental in understanding the local behavior of the curve around the point (9, 16). The problem combines the algebraic manipulation of implicit differentiation with the geometric interpretation of the derivative as the slope of a tangent line. This interplay between algebra and geometry is a recurring theme in calculus and underscores the power of calculus in analyzing curves and functions. Understanding the problem statement thoroughly is the first step in solving it effectively. We need to leverage the techniques of implicit differentiation, the chain rule, and the given condition to arrive at the final answer for y'(9).

Applying Implicit Differentiation

Now, let's embark on the process of applying implicit differentiation to the equation $\sqrt{x} + \sqrt{y} = 7$ . Our first step is to differentiate both sides of the equation with respect to x. Remember, we're treating y as a function of x, which is crucial for the correct application of the chain rule. The derivative of $\sqrt{x}$ with respect to x is (1/2)x^(-1/2), which simplifies to 1/(2 $\sqrt{x}$ ). For the term $\sqrt{y}$ , we need to employ the chain rule. We first differentiate $\sqrt{y}$ with respect to y, which gives us (1/2)y^(-1/2) or 1/(2 $\sqrt{y}$ ). Then, we multiply this by dy/dx because y is a function of x. This application of the chain rule is the hallmark of implicit differentiation and is essential for handling implicitly defined functions.

The derivative of the constant 7 on the right side of the equation is simply 0. Constants always have a derivative of zero because they do not change with respect to any variable. Putting it all together, the differentiated equation looks like this: 1/(2 $\sqrt{x}$ ) + [1/(2 $\sqrt{y}$ )] * (dy/dx) = 0. Our next step is to isolate dy/dx, which represents the derivative we are trying to find. To do this, we first subtract 1/(2 $\sqrt{x}$ ) from both sides of the equation, and then multiply both sides by 2 $\sqrt{y}$ . This gives us dy/dx = - $\sqrt{y}$ / $\sqrt{x}$ . This equation provides a general expression for the derivative dy/dx in terms of x and y. It tells us how the rate of change of y with respect to x varies depending on the values of x and y. We have successfully navigated the core steps of implicit differentiation, transforming the original equation into one that expresses the derivative dy/dx. The next step is to use the given condition y(9) = 16 to find the specific value of y'(9).

Evaluating $y'(9)$

Having found the general expression for dy/dx as - $\sqrt{y}$ / $\sqrt{x}$ , we now proceed to evaluate y'(9). This means we want to find the value of the derivative at the specific point where x = 9. Recall that we were given the condition y(9) = 16. This provides us with the corresponding value of y when x is 9. This is crucial because our expression for dy/dx is in terms of both x and y. We need the value of y to plug into the expression along with x = 9. Substituting x = 9 and y = 16 into the expression dy/dx = - $\sqrt{y}$ / $\sqrt{x}$ , we get dy/dx = - $\sqrt{16}$ / $\sqrt{9}$ .

Now, we simplify the expression. The square root of 16 is 4, and the square root of 9 is 3. Therefore, dy/dx = -4/3. This means that at the point (9, 16) on the curve defined by the equation $\sqrt{x} + \sqrt{y} = 7$ , the slope of the tangent line is -4/3. In other words, at this point, y is decreasing at a rate of 4/3 units for every 1 unit increase in x. This result provides a precise quantitative description of the rate of change of y with respect to x at a specific point. The negative sign indicates that y is decreasing as x increases, which is consistent with the shape of the curve defined by the original equation. The magnitude of the slope, 4/3, gives us the steepness of the tangent line at that point. We have successfully evaluated y'(9) by combining the general expression for dy/dx obtained through implicit differentiation with the specific condition y(9) = 16. This demonstrates the power of calculus in analyzing the behavior of functions at specific points.

Conclusion: Finding $y'(9) = -4/3$

In conclusion, we have successfully navigated the process of finding y'(9) for the equation $\sqrt{x} + \sqrt{y} = 7$ given the condition y(9) = 16. We began by understanding the concept of implicit differentiation, which is a technique used to find the derivative of implicitly defined functions. We then applied this technique to our equation, differentiating both sides with respect to x and using the chain rule appropriately. This led us to a general expression for dy/dx in terms of x and y: dy/dx = - $\sqrt{y}$ / $\sqrt{x}$ .

Next, we utilized the given condition, y(9) = 16, to evaluate y'(9). By substituting x = 9 and y = 16 into our expression for dy/dx, we found that y'(9) = -4/3. This result tells us that at the point (9, 16) on the curve defined by the equation $\sqrt{x} + \sqrt{y} = 7$ , the slope of the tangent line is -4/3. This means that y is decreasing at a rate of 4/3 units for every 1 unit increase in x at this specific point. This problem demonstrates the power and utility of implicit differentiation in analyzing the behavior of implicitly defined functions. It highlights the interplay between algebraic manipulation and geometric interpretation in calculus. The process of finding y'(9) involves not only the mechanical steps of differentiation but also the conceptual understanding of what the derivative represents – the instantaneous rate of change and the slope of the tangent line. This problem serves as a valuable example of how calculus can be used to solve problems involving related rates and the behavior of curves.