Implicit Differentiation Dy/dx For 5y^2 = (2x-3)/(2x+3)

July 12, 2025 by Scholario Team 56 views

Implicit Differentiation Techniques for Solving Complex Equations

In the realm of calculus, implicit differentiation stands as a powerful technique for finding the derivative of a function, especially when the function is not explicitly defined in the form y = f(x). This method is particularly useful when dealing with equations where it's difficult or impossible to isolate one variable in terms of the other. In this comprehensive guide, we'll delve into the intricacies of implicit differentiation, demonstrating its application with a detailed walkthrough of the equation 5y^2 = (2x-3)/(2x+3). This exploration will not only solidify your understanding of the technique but also equip you with the skills to tackle a wide array of calculus problems.

Understanding Implicit Differentiation

Implicit differentiation is a method used when we have an equation that relates x and y, but y is not explicitly given as a function of x. In other words, we can't easily isolate y on one side of the equation. This often happens in more complex equations, and that's where implicit differentiation shines. 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 is where the chain rule comes into play, as we'll see in our example. When differentiating terms involving y, we must remember to apply the chain rule, which introduces a dy/dx term. This dy/dx represents the derivative of y with respect to x, which is precisely what we aim to find. By carefully applying the chain rule and algebraic manipulation, we can isolate dy/dx and obtain an expression for the derivative. This technique opens doors to solving a variety of problems, including finding tangent lines to curves, related rates problems, and analyzing the behavior of implicitly defined functions.

The Power of Implicit Differentiation in Calculus

Calculus often presents us with equations that aren't straightforward. Sometimes, we encounter equations where it's either very difficult or outright impossible to isolate y and express it explicitly as a function of x. This is where implicit differentiation becomes an indispensable tool. Instead of needing to rewrite the equation, we can directly differentiate both sides with respect to x. This technique elegantly handles situations where y is intertwined with x in a complex manner. The beauty of implicit differentiation lies in its ability to provide us with dy/dx even when we don't have an explicit formula for y in terms of x. This is particularly useful in advanced calculus topics like related rates and optimization problems. Moreover, implicit differentiation allows us to analyze the behavior of implicitly defined functions, which are prevalent in various fields of mathematics, physics, and engineering. By mastering implicit differentiation, you gain a powerful technique for tackling a wider range of calculus problems and deepen your understanding of mathematical relationships.

Step-by-Step Solution for 5y^2 = (2x-3)/(2x+3)

Let's dive into solving the given equation: 5y^2 = (2x-3)/(2x+3). Our goal is to find dy/dx using implicit differentiation. This involves a series of steps, each building upon the previous one, to arrive at the desired derivative. We'll start by differentiating both sides of the equation with respect to x, carefully applying the chain rule where necessary. Then, we'll use algebraic manipulation to isolate dy/dx, revealing the relationship between the rates of change of x and y. This step-by-step approach will not only provide the solution but also illustrate the thought process behind implicit differentiation, making it easier to apply the technique to other problems.

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 a crucial step as it sets the stage for applying the chain rule and isolating dy/dx. When differentiating, remember that y is a function of x, so we'll need to use the chain rule whenever we encounter a term involving y. On the left side, we have 5y^2. Differentiating this with respect to x gives us 10y(dy/dx), applying the power rule and the chain rule. On the right side, we have (2x-3)/(2x+3), which is a quotient. We'll need to use the quotient rule to differentiate this. The quotient rule states that the derivative of u/v is (v(du/dx) - u(dv/dx)) / v^2. By carefully applying these differentiation rules to both sides, we'll create an equation that includes dy/dx, which we can then solve for.

So, differentiating both sides of 5y^2 = (2x-3)/(2x+3) with respect to x, we get:

d/dx [5y^2] = d/dx [(2x-3)/(2x+3)]

Applying the power rule and chain rule on the left side:

10y (dy/dx) = d/dx [(2x-3)/(2x+3)]

2. Apply the Quotient Rule

On the right side of the equation, we have a quotient, which requires the quotient rule for differentiation. The quotient rule is a fundamental tool in calculus for finding the derivative of a fraction where both the numerator and denominator are functions of x. It states that if we have a function of the form u(x) / v(x), its derivative is given by (v(x)u'(x) - u(x)v'(x)) / [v(x)]^2. Applying this rule correctly is crucial for obtaining the correct derivative. In our case, u(x) = 2x - 3 and v(x) = 2x + 3. We need to find the derivatives of u(x) and v(x) individually before plugging them into the quotient rule formula. By carefully applying the quotient rule, we'll transform the right side of the equation into a more manageable form, paving the way for isolating dy/dx.

Applying the quotient rule to the right side, where u = 2x - 3 and v = 2x + 3:

d/dx [(2x-3)/(2x+3)] = [(2x+3)(2) - (2x-3)(2)] / (2x+3)^2

Simplifying the numerator:

= [4x + 6 - 4x + 6] / (2x+3)^2
= 12 / (2x+3)^2

3. Substitute Back into the Equation

Now that we've differentiated both sides of the original equation, we need to substitute the derivative of the right side back into our equation. This step brings together the results of our previous differentiation steps and sets us up to isolate dy/dx. We've found that the derivative of 5y^2 with respect to x is 10y(dy/dx), and the derivative of (2x-3)/(2x+3) with respect to x is 12 / (2x+3)^2. Substituting these back into our equation, we get 10y(dy/dx) = 12 / (2x+3)^2. This equation now explicitly relates dy/dx to x and y. The next step involves algebraic manipulation to isolate dy/dx on one side of the equation.

Substituting this back into our equation, we get:

10y (dy/dx) = 12 / (2x+3)^2

4. Isolate dy/dx

The final step in finding dy/dx is to isolate it on one side of the equation. This involves using algebraic manipulation to get dy/dx by itself. In our case, we have the equation 10y(dy/dx) = 12 / (2x+3)^2. To isolate dy/dx, we need to divide both sides of the equation by 10y. This will give us an expression for dy/dx in terms of x and y. This resulting expression represents the derivative of y with respect to x, which tells us the instantaneous rate of change of y as x changes. It's important to simplify the expression as much as possible to obtain the most concise form of the derivative. This final step completes the process of implicit differentiation, providing us with the desired dy/dx.

To isolate dy/dx, divide both sides by 10y:

dy/dx = [12 / (2x+3)^2] / 10y
dy/dx = 12 / [10y(2x+3)^2]
dy/dx = 6 / [5y(2x+3)^2]

Final Answer

Therefore, the derivative dy/dx for the equation 5y^2 = (2x-3)/(2x+3) is:

dy/dx = 6 / [5y(2x+3)^2]

This expression gives the rate of change of y with respect to x for any point (x, y) on the curve defined by the original equation. This result showcases the power of implicit differentiation in finding derivatives for functions that are not explicitly defined.

Conclusion: Mastering Implicit Differentiation

In conclusion, implicit differentiation is a vital technique in calculus, allowing us to find derivatives of functions that are not explicitly defined. By differentiating both sides of an equation with respect to x and applying the chain rule, we can solve for dy/dx. Our step-by-step solution for the equation 5y^2 = (2x-3)/(2x+3) demonstrates the process, from differentiating both sides and applying the quotient rule to isolating dy/dx. Mastering implicit differentiation expands your calculus toolkit, enabling you to tackle more complex problems and gain a deeper understanding of mathematical relationships. With practice and careful attention to detail, you can confidently apply this technique to a wide range of calculus challenges.