Implicit Differentiation Solving (7xy+6)^2=14y For Dy/dx

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Introduction to Implicit Differentiation

In the realm of calculus, implicit differentiation stands as a powerful technique for finding the derivative of a function, especially when that function is not explicitly defined in the form y = f(x). Instead, we encounter equations where x and y are intertwined, making it challenging to isolate y. This is where implicit differentiation shines, allowing us to find dy/dx even when y is not explicitly expressed as a function of x. To begin, it's important to understand what implicit differentiation really means. Often, in standard calculus problems, we deal with explicit functions, where y is clearly defined in terms of x, such as y = x^2 + 3x - 5. However, many equations in mathematics and its applications do not present themselves so neatly. Consider equations like x^2 + y^2 = 25 (a circle) or the one we're tackling today, (7xy + 6)^2 = 14y. In these cases, y is implicitly defined as a function of x. This means that y is a function of x, but it's not explicitly isolated on one side of the equation. To tackle these implicit equations, we employ a method that directly differentiates both sides of the equation with respect to x, keeping in mind that y is a function of x. This approach cleverly uses the chain rule, a cornerstone of differentiation, to handle terms involving y. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. This becomes particularly important when we differentiate terms involving y because we must account for the fact that y is changing with respect to x. In simpler terms, we treat y as a function within a function.

Understanding the chain rule is paramount to mastering implicit differentiation. When we differentiate an expression like y^2 with respect to x, we don't simply get 2y. Instead, we apply the chain rule, which gives us 2y(dy/dx)*. This is because we're differentiating y^2 with respect to y, and then multiplying by the derivative of y with respect to x. This dy/dx term is the key to unlocking the derivative in implicit functions. By differentiating every term in the equation this way, we create an equation that includes dy/dx. Then, it becomes a matter of algebraic manipulation to isolate dy/dx on one side of the equation, thus finding the derivative we seek. In the context of our specific problem, (7xy + 6)^2 = 14y, implicit differentiation is the perfect tool. We could try to solve this equation explicitly for y, but that would likely be a complex and messy process. Implicit differentiation offers a more direct and elegant approach. It allows us to work with the equation as it is, differentiating both sides and using the chain rule to handle the y terms. This method not only simplifies the process but also highlights the interconnectedness of x and y in the equation, providing a deeper understanding of their relationship.

Problem Setup: (7xy + 6)^2 = 14y

Let's dive into the specifics of our problem: (7xy + 6)^2 = 14y. This equation presents a classic scenario for implicit differentiation. We have x and y entangled in a way that makes it difficult to directly express y as a function of x. The presence of the squared term (7xy + 6)^2 and the product 7xy further complicate matters, necessitating a careful application of the chain rule and the product rule during differentiation. To successfully tackle this problem, we need to meticulously differentiate both sides of the equation with respect to x. This involves recognizing that y is a function of x and applying the chain rule whenever we encounter a y term. The left side of the equation, (7xy + 6)^2, requires a combination of the chain rule and the product rule. The chain rule is needed because we have a function squared, and the product rule is needed because the function inside the parentheses, 7xy, is a product of two functions, 7x and y. On the right side of the equation, 14y, we simply apply the chain rule, as we are differentiating a constant multiple of y with respect to x.

Before we jump into the differentiation, it's beneficial to outline the steps we'll take. This will provide a clear roadmap for our solution and help us avoid common pitfalls. First, we'll differentiate both sides of the equation with respect to x, carefully applying the chain rule and the product rule as needed. Second, we'll simplify the resulting equation, combining like terms and rearranging to isolate dy/dx. This step often involves algebraic manipulation, such as factoring and dividing. Finally, we'll solve for dy/dx, expressing it in terms of x and y. This will give us the derivative of y with respect to x for the given implicit equation. It's crucial to pay close attention to the order of operations and to double-check our work at each step. Implicit differentiation can be prone to errors if the chain rule or product rule is misapplied. By taking our time and being methodical, we can ensure that we arrive at the correct solution. In the next section, we'll begin the differentiation process, breaking it down step by step to make it clear and understandable. We'll see how the chain rule and product rule come into play and how we can carefully apply them to this equation. The goal is not just to find the answer, but to understand the process and the underlying principles of implicit differentiation.

Step-by-Step Differentiation

Now, let's embark on the core of the problem: differentiating both sides of the equation (7xy + 6)^2 = 14y with respect to x. This is where the magic of implicit differentiation happens, and where careful application of the chain rule and product rule is paramount. We'll start with the left side of the equation, (7xy + 6)^2. This term is a composite function, meaning we have a function inside another function. The outer function is the squaring function, and the inner function is 7xy + 6. To differentiate this, we'll apply the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. So, the derivative of (7xy + 6)^2 with respect to x is 2(7xy + 6) multiplied by the derivative of (7xy + 6) with respect to x. Now, we need to find the derivative of the inner function, 7xy + 6, with respect to x. This is where the product rule comes into play. The product rule states that the derivative of the product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function. In our case, we have the product 7x and y. The derivative of 7x with respect to x is simply 7. The derivative of y with respect to x is dy/dx, which is what we're trying to find. Applying the product rule, the derivative of 7xy with respect to x is 7y + 7x(dy/dx). The derivative of the constant 6 with respect to x is 0. Putting it all together, the derivative of (7xy + 6) with respect to x is 7y + 7x(dy/dx) + 0, which simplifies to 7y + 7x(dy/dx). Going back to our original application of the chain rule, the derivative of (7xy + 6)^2 with respect to x is 2(7xy + 6) * [7y + 7x(dy/dx)]. This is a significant step, as we've successfully differentiated the left side of the equation using both the chain rule and the product rule. Next, we'll tackle the right side of the equation, 14y. This term is simpler to differentiate, but we still need to apply the chain rule because y is a function of x. The derivative of 14y with respect to x is simply 14(dy/dx). Now, we have the derivatives of both sides of the equation. The derivative of the left side is 2(7xy + 6) * [7y + 7x(dy/dx)], and the derivative of the right side is 14(dy/dx). We can now set these equal to each other, giving us the equation 2(7xy + 6) * [7y + 7x(dy/dx)] = 14(dy/dx). This equation contains dy/dx, which is what we want to solve for. In the next section, we'll focus on simplifying this equation and isolating dy/dx. This will involve algebraic manipulation, such as distributing, combining like terms, and factoring.

Isolating dy/dx

With the differentiation complete, our next crucial step is isolating dy/dx in the equation 2(7xy + 6) * [7y + 7x(dy/dx)] = 14(dy/dx). This involves a series of algebraic manipulations to group all terms containing dy/dx on one side of the equation and all other terms on the other side. The first step in this process is to distribute the terms on the left side of the equation. We have 2(7xy + 6) multiplying the expression [7y + 7x(dy/dx)]. Distributing this, we get: 2(7xy + 6) * 7y + 2(7xy + 6) * 7x(dy/dx) = 14(7xy + 6)y + 14x(7xy + 6)(dy/dx). Now our equation looks like this: 14y(7xy + 6) + 14x(7xy + 6)(dy/dx) = 14(dy/dx). This expansion reveals the terms containing dy/dx more clearly. The next step is to move all terms containing dy/dx to one side of the equation and all other terms to the other side. To do this, we'll subtract 14x(7xy + 6)(dy/dx) from both sides of the equation, resulting in: 14y(7xy + 6) = 14(dy/dx) - 14x(7xy + 6)(dy/dx). Now, all terms containing dy/dx are on the right side of the equation, and all other terms are on the left side. The next step is to factor out dy/dx from the right side of the equation. We have two terms on the right side, both of which contain dy/dx. Factoring out dy/dx, we get: 14y(7xy + 6) = dy/dx * [14 - 14x(7xy + 6)]. Now, dy/dx is isolated as a factor on the right side of the equation. Finally, to solve for dy/dx, we need to divide both sides of the equation by the expression in the brackets: [14 - 14x(7xy + 6)]. This gives us: dy/dx = [14y(7xy + 6)] / [14 - 14x(7xy + 6)]. We have now successfully isolated dy/dx. However, our expression can be simplified further. Notice that both the numerator and the denominator have a common factor of 14. We can divide both the numerator and the denominator by 14 to simplify the expression. This gives us: dy/dx = [y(7xy + 6)] / [1 - x(7xy + 6)]. This is a more simplified expression for dy/dx. We can further expand the terms in the numerator and the denominator to get: dy/dx = (7xy^2 + 6y) / (1 - 7x^2y - 6x). This is the final expression for dy/dx, expressed in terms of x and y.

Final Solution and Interpretation

After meticulously applying implicit differentiation and algebraic manipulation, we arrive at the final solution for dy/dx for the equation (7xy + 6)^2 = 14y: dy/dx = (7xy^2 + 6y) / (1 - 7x^2y - 6x). This expression represents the derivative of y with respect to x, even though y is not explicitly defined as a function of x. The beauty of implicit differentiation lies in its ability to handle such situations, providing us with a powerful tool for analyzing relationships between variables. Now that we have the solution, it's crucial to interpret its meaning and understand what this derivative tells us. The derivative dy/dx represents the instantaneous rate of change of y with respect to x. In geometric terms, it gives us the slope of the tangent line to the curve defined by the equation (7xy + 6)^2 = 14y at any point (x, y) on the curve. This is a fundamental concept in calculus, with applications in various fields, including physics, engineering, and economics. To fully grasp the implications of this derivative, let's consider some specific scenarios. Suppose we want to find the slope of the tangent line at a particular point on the curve. We would simply substitute the x and y coordinates of that point into our expression for dy/dx. The resulting value would be the slope of the tangent line at that point. This allows us to analyze the behavior of the curve at different locations. For instance, if dy/dx is positive at a certain point, it means that y is increasing as x increases at that point. Conversely, if dy/dx is negative, it means that y is decreasing as x increases. If dy/dx is zero, it means that the tangent line is horizontal, indicating a local maximum or minimum point. Furthermore, the expression for dy/dx itself provides insights into the relationship between x and y. The complexity of the expression reflects the complexity of the original equation. The presence of both x and y in the derivative highlights the implicit nature of the relationship between the variables. This means that the rate of change of y with respect to x depends on both the x and y values at a particular point. In conclusion, implicit differentiation has allowed us to find dy/dx for the equation (7xy + 6)^2 = 14y. Our final solution, dy/dx = (7xy^2 + 6y) / (1 - 7x^2y - 6x), provides a powerful tool for analyzing the behavior of the curve defined by this equation. By understanding the meaning of the derivative and its implications, we can gain valuable insights into the relationship between x and y.

Conclusion

In this comprehensive exploration, we've successfully navigated the intricacies of implicit differentiation to find dy/dx for the equation (7xy + 6)^2 = 14y. This journey has not only provided us with a solution but has also illuminated the fundamental principles and applications of this powerful calculus technique. We began by understanding the essence of implicit differentiation, recognizing its significance in scenarios where y is not explicitly defined as a function of x. We contrasted it with explicit differentiation, highlighting the unique challenges and advantages of the implicit approach. We emphasized the crucial role of the chain rule in handling terms involving y, as well as the product rule when dealing with products of functions. The problem setup involved carefully analyzing the given equation, (7xy + 6)^2 = 14y, and identifying the need for implicit differentiation due to the intertwined nature of x and y. We outlined a step-by-step approach, emphasizing the importance of methodical differentiation and algebraic manipulation. The differentiation process itself was a meticulous application of the chain rule and the product rule. We broke down each step, ensuring clarity and accuracy. We first differentiated the left side of the equation, (7xy + 6)^2, using the chain rule and the product rule. Then, we differentiated the right side of the equation, 14y, using the chain rule. We carefully combined these derivatives to form an equation containing dy/dx. The subsequent task of isolating dy/dx involved a series of algebraic manipulations. We distributed terms, moved terms containing dy/dx to one side, factored out dy/dx, and finally solved for dy/dx. This process showcased the importance of algebraic skills in conjunction with calculus techniques. Our final solution, dy/dx = (7xy^2 + 6y) / (1 - 7x^2y - 6x), represents the derivative of y with respect to x for the given equation. We emphasized the importance of interpreting this derivative, recognizing it as the instantaneous rate of change of y with respect to x and the slope of the tangent line to the curve at any point. We discussed how to use this derivative to analyze the behavior of the curve and gain insights into the relationship between x and y. In essence, this exploration has demonstrated the power and versatility of implicit differentiation. It's a technique that allows us to tackle equations that would be difficult or impossible to solve using explicit methods. By mastering implicit differentiation, we expand our calculus toolkit and gain a deeper understanding of mathematical relationships. This understanding has far-reaching implications in various fields, making implicit differentiation a valuable skill for anyone pursuing studies in mathematics, science, engineering, or economics. The journey through this problem serves as a testament to the beauty and elegance of calculus, where seemingly complex equations can be unravelled through careful application of fundamental principles.