Finding Dy/dx For Xy³ - 2y² + X²y - 8 = 0 A Step-by-Step Guide
Implicit differentiation is a powerful technique in calculus that allows us to find the derivative of a function even when it is not explicitly defined in the form y = f(x). This article delves into the process of finding dy/dx for the equation xy³ - 2y² + x²y - 8 = 0 using implicit differentiation. We will break down each step, providing a comprehensive explanation to ensure a clear understanding of the method. Mastering implicit differentiation is crucial for solving various problems in calculus and related fields, making this a fundamental concept for students and professionals alike.
Understanding Implicit Differentiation
Before diving into the specifics of the given equation, let's establish a solid understanding of what implicit differentiation entails. In explicit functions, y is expressed directly in terms of x, such as y = x² + 3x - 1. However, in implicit functions, the relationship between x and y is defined implicitly through an equation, where y is not isolated on one side. The equation xy³ - 2y² + x²y - 8 = 0 is a perfect example of such an implicit function. Implicit differentiation allows us to find the derivative dy/dx without the need to explicitly solve for y. The key principle behind this method is the application of the chain rule. When differentiating terms involving y with respect to x, we must remember that y is itself a function of x, and thus, we need to multiply by dy/dx. This technique is essential for handling equations where isolating y is difficult or impossible, making it a versatile tool in calculus.
The Chain Rule in Implicit Differentiation
The chain rule is a fundamental concept in calculus that dictates how to differentiate composite functions. A composite function is a function that is composed of another function, such as f(g(x)). 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. Mathematically, this is expressed as:
d/dx [f(g(x))] = f'(g(x)) * g'(x)
In the context of implicit differentiation, the chain rule is particularly important when differentiating terms involving y with respect to x. Since y is an implicit function of x, we treat it as y(x). When we differentiate a term like y² with respect to x, we apply the chain rule as follows:
d/dx (y²) = 2y * dy/dx
Here, the outer function is the squaring function, and the inner function is y(x). We first differentiate the outer function with respect to y, which gives us 2y. Then, we multiply by the derivative of the inner function, which is dy/dx. This process ensures that we account for the fact that y is changing with respect to x. Understanding and correctly applying the chain rule is crucial for mastering implicit differentiation and accurately finding derivatives of implicit functions.
Product Rule and Quotient Rule
In addition to the chain rule, the product rule and quotient rule are essential tools when performing implicit differentiation. These rules help us differentiate functions that are products or quotients of other functions. 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. Mathematically, if we have two functions u(x) and v(x), their product's derivative is:
d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
The quotient rule, on the other hand, is used to differentiate the quotient of two functions. It states that the derivative of the quotient of two functions is the derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator. If we have two functions u(x) and v(x), their quotient's derivative is:
d/dx [u(x)/v(x)] = [u'(x)v(x) - u(x)v'(x)] / [v(x)]²
In the context of implicit differentiation, these rules are particularly useful when dealing with terms that involve both x and y. For example, in the equation xy³ - 2y² + x²y - 8 = 0, we encounter terms like xy³ and x²y, which require the product rule to differentiate correctly. Mastering these rules is crucial for accurately applying implicit differentiation to a wide range of equations.
Step-by-Step Solution
Now, let's apply implicit differentiation to the equation xy³ - 2y² + x²y - 8 = 0. We will proceed step-by-step to ensure clarity and accuracy.
Step 1: Differentiate Both Sides with Respect to x
The first step in implicit differentiation is to differentiate both sides of the equation with respect to x. This is crucial for maintaining the equality of the equation. Applying the differentiation operator d/dx to both sides of the equation xy³ - 2y² + x²y - 8 = 0, we get:
d/dx (xy³ - 2y² + x²y - 8) = d/dx (0)
Since the derivative of a constant is zero, the right side of the equation becomes 0. Now, we need to differentiate the left side term by term. This involves applying the sum/difference rule of differentiation, which states that the derivative of a sum or difference of terms is the sum or difference of their derivatives. Thus, we have:
d/dx (xy³) - d/dx (2y²) + d/dx (x²y) - d/dx (8) = 0
This step sets the stage for the subsequent application of the product rule and chain rule, which are essential for differentiating the individual terms involving both x and y. By differentiating both sides, we ensure that we are accounting for the relationship between x and y as they change together.
Step 2: Apply the Product Rule and Chain Rule
This is the core step where we apply the product rule and chain rule to differentiate each term on the left side of the equation. Let's break down each term individually:
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d/dx (xy³): This term requires the product rule. Let u = x and v = y³. Then, du/dx = 1 and dv/dx = 3y² * dy/dx (using the chain rule). Applying the product rule, we get:
d/dx (xy³) = (1)(y³) + (x)(3y² * dy/dx) = y³ + 3xy² * dy/dx
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d/dx (2y²): This term requires the chain rule. The derivative of 2y² with respect to y is 4y. Applying the chain rule, we get:
d/dx (2y²) = 4y * dy/dx
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d/dx (x²y): This term also requires the product rule. Let u = x² and v = y. Then, du/dx = 2x and dv/dx = dy/dx. Applying the product rule, we get:
d/dx (x²y) = (2x)(y) + (x²)(dy/dx) = 2xy + x² * dy/dx
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d/dx (8): The derivative of a constant is zero, so:
d/dx (8) = 0
Now, we substitute these results back into the equation from Step 1:
(y³ + 3xy² * dy/dx) - (4y * dy/dx) + (2xy + x² * dy/dx) - 0 = 0
This step is critical as it correctly applies the necessary differentiation rules to each term, setting up the equation for isolating dy/dx in the next step.
Step 3: Rearrange the Equation to Isolate dy/dx
Now that we have differentiated each term, the next step is to rearrange the equation to isolate dy/dx. This involves grouping all the terms containing dy/dx on one side of the equation and moving all other terms to the other side. From the previous step, we have:
y³ + 3xy² * dy/dx - 4y * dy/dx + 2xy + x² * dy/dx = 0
First, let's move the terms without dy/dx to the right side of the equation:
3xy² * dy/dx - 4y * dy/dx + x² * dy/dx = -y³ - 2xy
Next, we factor out dy/dx from the terms on the left side:
dy/dx (3xy² - 4y + x²) = -y³ - 2xy
This rearrangement simplifies the equation and makes it clear that we can now isolate dy/dx by dividing both sides by the expression in the parentheses. This step is a crucial algebraic manipulation that brings us closer to finding the derivative dy/dx.
Step 4: Solve for dy/dx
Finally, we solve for dy/dx by dividing both sides of the equation by the expression (3xy² - 4y + x²). From the previous step, we have:
dy/dx (3xy² - 4y + x²) = -y³ - 2xy
Dividing both sides by (3xy² - 4y + x²), we get:
dy/dx = (-y³ - 2xy) / (3xy² - 4y + x²)
This is the derivative of y with respect to x for the given implicit equation. The result expresses dy/dx in terms of both x and y, which is typical for implicit differentiation. This final step completes the process of finding the derivative, providing a clear and concise expression for dy/dx. It's important to note that the derivative is a function of both x and y, reflecting the implicit relationship between the two variables.
Final Answer
Therefore, for the equation xy³ - 2y² + x²y - 8 = 0, the derivative dy/dx is:
dy/dx = (-y³ - 2xy) / (3xy² - 4y + x²)
This result provides the rate of change of y with respect to x at any point (x, y) that satisfies the original equation. Understanding and being able to apply implicit differentiation is a valuable skill in calculus, allowing us to solve problems involving implicitly defined functions effectively. This technique is not only applicable in mathematics but also in various fields such as physics, engineering, and economics, where implicit relationships between variables are common. Mastering implicit differentiation enhances one's problem-solving capabilities and provides a deeper understanding of the relationships between variables in complex systems.
