Implicit Differentiation Find Dy/dx For 6xy + Y² = 7x + Y
Introduction to Implicit Differentiation
In the realm of calculus, implicit differentiation stands as a powerful technique for finding the derivative of a function, particularly when that function is not explicitly defined in the form y = f(x). Instead, we encounter equations where x and y are intertwined, like our example: 6xy + y² = 7x + y. This equation implicitly defines y as a function of x, and to unravel the rate of change, dy/dx, we turn to implicit differentiation.
At its core, implicit differentiation leverages the chain rule, a fundamental concept in calculus that governs the differentiation of composite functions. When we differentiate terms involving y with respect to x, we must remember that y is itself a function of x. This means we differentiate y as usual but then multiply by dy/dx to account for the chain rule. Ignoring this crucial step would lead to incorrect results. This is because the chain rule is essential for understanding how the rate of change of an outer function is affected by the rate of change of its inner function. Think of it as a ripple effect: a small change in x affects y, and that change in y in turn affects the entire expression. Implicit differentiation allows us to capture this ripple effect accurately.
The beauty of implicit differentiation lies in its ability to handle equations that would be incredibly difficult or even impossible to solve for y explicitly. Imagine trying to isolate y in our equation 6xy + y² = 7x + y. The algebraic manipulations would be cumbersome and might not even lead to a closed-form solution. Implicit differentiation bypasses this hurdle, allowing us to find dy/dx directly from the implicit equation. This makes it an indispensable tool for dealing with complex relationships between variables in various fields, including physics, engineering, and economics. By understanding and mastering this technique, you unlock a powerful way to analyze and model real-world phenomena where variables are interconnected in intricate ways.
Step-by-Step Solution: Differentiating 6xy + y² = 7x + y
Let's dive into the step-by-step process of finding dy/dx for the equation 6xy + y² = 7x + y using implicit differentiation. This process involves a careful application of differentiation rules and algebraic manipulation. We will break down each step in detail to ensure clarity and understanding.
1. Differentiate both sides of the equation with respect to x:
The cornerstone of implicit differentiation is applying the derivative operator, d/dx, to both sides of the equation. This maintains the equality and sets the stage for unraveling dy/dx. Remember, we are treating y as a function of x, so the chain rule will come into play whenever we differentiate terms involving y. This step is crucial because it transforms the implicit equation into a form where we can isolate the derivative we are looking for. By differentiating both sides, we are essentially capturing the instantaneous rate of change of both sides of the equation with respect to x. This is the foundation of finding how y changes as x changes.
Applying d/dx to 6xy + y² = 7x + y yields:
d/dx (6xy + y²) = d/dx (7x + y)
2. Apply the product rule and chain rule:
Now comes the crucial step of applying the appropriate differentiation rules. On the left-hand side, we encounter the term 6xy, which necessitates the product rule. The product rule states that the derivative of a product of two functions, u(x) and v(x), is given by (uv)' = u'v + uv'. In our case, u(x) = 6x and v(x) = y. The derivative of 6x with respect to x is simply 6. The derivative of y with respect to x is dy/dx. Applying the product rule diligently is essential for accurately capturing the interaction between x and y in the equation. Failing to do so will lead to an incorrect derivative.
For the term y², we employ the chain rule. The chain rule is invoked whenever we differentiate a composite function, that is, a function within a function. Here, we have the square function applied to y, where y is itself a function of x. The chain rule states that the derivative of f(g(x)) is f'(g(x)) * g'(x). In our case, the derivative of y² with respect to y is 2y, and then we multiply by dy/dx because y is a function of x. This multiplication by dy/dx is the hallmark of implicit differentiation and a critical step to remember.
On the right-hand side, the derivative of 7x with respect to x is simply 7. The derivative of y with respect to x is, as always, dy/dx. Putting it all together, we get:
6(x * dy/dx + y * 1) + 2y(dy/dx) = 7 + dy/dx
This simplifies to:
6y + 6x(dy/dx) + 2y(dy/dx) = 7 + dy/dx
3. Collect terms involving dy/dx:
The next step is to strategically rearrange the equation to isolate all terms containing dy/dx on one side. This is a standard algebraic technique used to solve for a specific variable. We move all terms with dy/dx to the left-hand side and all other terms to the right-hand side. This grouping is essential for the subsequent factoring step, which will allow us to finally solve for dy/dx. By collecting like terms, we are essentially organizing the equation in a way that makes the target variable, dy/dx, stand out and become the subject of the equation.
Subtracting dy/dx from both sides and subtracting 6y from both sides, we get:
6x(dy/dx) + 2y(dy/dx) - dy/dx = 7 - 6y
4. Factor out dy/dx:
Now, with all the dy/dx terms on one side, we can factor it out. This is a crucial algebraic manipulation that allows us to treat dy/dx as a single variable and isolate it. Factoring out dy/dx is like extracting a common factor from a polynomial expression. It simplifies the equation and brings us closer to our goal of finding an expression for dy/dx. This step demonstrates the power of algebraic techniques in solving calculus problems.
Factoring out dy/dx gives us:
(dy/dx)(6x + 2y - 1) = 7 - 6y
5. Solve for dy/dx:
Finally, we arrive at the last step: solving for dy/dx. This involves dividing both sides of the equation by the expression multiplying dy/dx. This isolates dy/dx on the left-hand side, giving us an explicit formula for the derivative. This step is the culmination of all the previous steps, and it provides the answer we have been seeking: the derivative of y with respect to x.
Dividing both sides by (6x + 2y - 1), we obtain:
dy/dx = (7 - 6y) / (6x + 2y - 1)
Conclusion
We have successfully used implicit differentiation to find dy/dx for the equation 6xy + y² = 7x + y. The result, dy/dx = (7 - 6y) / (6x + 2y - 1), expresses the derivative in terms of both x and y. This is a common characteristic of implicit differentiation, as the derivative often depends on both variables when the function is not explicitly defined. Understanding this method opens doors to solving a wide array of calculus problems where direct differentiation is not feasible. The key takeaways are the application of the chain rule and the algebraic manipulation to isolate dy/dx. Mastering these concepts will significantly enhance your calculus skills and your ability to tackle complex mathematical problems.
Further Practice
To solidify your understanding of implicit differentiation, try applying this technique to similar problems. Practice is key to mastering any mathematical concept, and implicit differentiation is no exception. Working through various examples will help you develop the intuition and skills needed to tackle more complex problems. Consider exploring other implicit equations and finding their derivatives. This will not only reinforce your understanding of the steps involved but also expose you to different scenarios and challenges. Remember, the more you practice, the more confident and proficient you will become in using implicit differentiation.
