Finding Dy/dx: A Step-by-Step Guide For $3y + 5x - Xy = 0$

Hey math enthusiasts! Today, we're diving into a classic calculus problem: finding $\frac{dy}{dx}$ when given an implicit equation. Specifically, we'll tackle the equation $3y + 5x - xy = 0$. Don't worry if this sounds a bit intimidating; we'll break it down into easy-to-follow steps. This is a fundamental concept in calculus, so understanding how to differentiate implicit equations is super important. We will explore how to approach this kind of problem methodically, ensuring that even if you're just starting, you can follow along. Understanding implicit differentiation unlocks a whole world of possibilities when working with equations that aren't explicitly solved for y. Let's get started, shall we?

Understanding the Problem and Implicit Differentiation

Before we jump into the solution, let's make sure we're all on the same page. What does it even mean to find $\frac{dy}{dx}$? This notation represents the derivative of y with respect to x. In simpler terms, it tells us how y changes as x changes. When we're given an equation like $3y + 5x - xy = 0$, where y isn't explicitly defined as a function of x, we use a technique called implicit differentiation. Implicit differentiation is like a superpower when dealing with equations where y is mixed up with x and it's tough (or impossible) to rearrange the equation to get y all by itself. We treat y as a function of x, even though we don't know the exact function. This means that whenever we differentiate a term involving y, we have to remember to multiply by $\frac{dy}{dx}$, thanks to the chain rule. The chain rule is our best friend here; it allows us to differentiate composite functions. Basically, if you have a function inside another function, the chain rule says you differentiate the outside, then multiply by the derivative of the inside. In this case, the 'inside' is y, and its derivative is $\frac{dy}{dx}$. So, the core idea is this: we're going to differentiate both sides of our equation with respect to x, keeping in mind that y is a function of x. Let's get ready to roll!

Differentiating the Equation Step-by-Step

Alright, guys, let's get down to the nitty-gritty and differentiate our equation: $3y + 5x - xy = 0$. We'll go term by term, carefully applying the rules of differentiation. This is where the fun begins! Remember, we want to find $\frac{dy}{dx}$, so our goal is to isolate this term. Here's how we do it:

1. Differentiate 3y: When we differentiate $3y$ with respect to x, we get $3\frac{dy}{dx}$. This is because we treat y as a function of x, so we need to multiply by its derivative.
2. Differentiate 5x: Differentiating $5x$ with respect to x gives us just $5$, since the derivative of x is 1.
3. Differentiate -xy: This is where it gets a little more interesting, because it involves the product of two functions of x: x and y. We need to use the product rule here. The product rule states that the derivative of $uv$ is $u'v + uv'$, where $u'$ and $v'$ are the derivatives of u and v, respectively. In our case, let $u = x$ and $v = y$. So, $u' = 1$ (the derivative of x with respect to x) and $v' = \frac{dy}{dx}$ (the derivative of y with respect to x). Applying the product rule to $-xy$, we get $-1*y - x\frac{dy}{dx}$, which simplifies to $-y - x\frac{dy}{dx}$.
4. Differentiate 0: The derivative of a constant (in this case, 0) is always 0.

So, putting it all together, our differentiated equation is: $3\frac{dy}{dx} + 5 - y - x\frac{dy}{dx} = 0$.

Isolating dy/dx

Now that we've differentiated the equation, our next mission is to isolate $\frac{dy}{dx}$. This means rearranging the equation so that all terms containing $\frac{dy}{dx}$ are on one side, and all other terms are on the other side. Here's how to do it:

1. Group terms with $\frac{dy}{dx}$: Move all the terms containing $\frac{dy}{dx}$ to one side of the equation and the other terms to the opposite side. Our equation is $3\frac{dy}{dx} + 5 - y - x\frac{dy}{dx} = 0$. Let's move the terms without $\frac{dy}{dx}$ to the right side and terms with $\frac{dy}{dx}$ to the left. We get $3\frac{dy}{dx} - x\frac{dy}{dx} = y - 5$.
2. Factor out $\frac{dy}{dx}$: Now, on the left side, we can factor out $\frac{dy}{dx}$. This gives us $\frac{dy}{dx}(3 - x) = y - 5$.
3. Solve for $\frac{dy}{dx}$: Finally, to solve for $\frac{dy}{dx}$, we divide both sides of the equation by $(3 - x)$. This isolates $\frac{dy}{dx}$ and gives us our final answer: $\frac{dy}{dx} = \frac{y - 5}{3 - x}$.

And there you have it! We've successfully found $\frac{dy}{dx}$ in terms of x and y. The answer is $\frac{y - 5}{3 - x}$.

The Final Answer and Understanding the Result

So, after all that work, the derivative $\frac{dy}{dx}$ for the equation $3y + 5x - xy = 0$ is $\frac{y - 5}{3 - x}$. What does this result actually mean? Well, this expression tells us the slope of the tangent line to the curve defined by the original equation at any point (x, y). The value of $\frac{dy}{dx}$ depends on both x and y, which is characteristic of implicit differentiation. Since y is not explicitly defined as a function of x, the slope varies with both variables. For any given point on the curve, you can plug in the x and y values into this expression to find the slope of the tangent line at that point. If you were to graph the original equation, you'd see a curve, and at each point on that curve, the value of $\frac{y - 5}{3 - x}$ gives you the slope. This is super useful for various applications, such as finding the rate of change of y with respect to x at a specific point, or for optimization problems where you might need to find where the slope is zero or some other specific value. This also demonstrates a fundamental principle of calculus: derivatives help us understand how things change, and with implicit differentiation, we can explore these changes even when the relationship between variables isn't straightforward. Understanding this concept is crucial for more advanced calculus topics. For instance, if you are studying related rates, this skill becomes indispensable. Also, it's a stepping stone to understanding partial derivatives and other important concepts. Keep practicing, and you'll become a pro at implicit differentiation in no time!

Summary and Key Takeaways

Let's recap what we've learned:

• Implicit Differentiation: This technique is used when y is not explicitly defined as a function of x. It's like a secret weapon for solving tricky equations.
• Chain Rule: Remember to multiply by $\frac{dy}{dx}$ whenever you differentiate a term involving y.
• Product Rule: Useful when differentiating terms like -xy.
• Step-by-Step Approach: Differentiate each term, group the terms, factor out $\frac{dy}{dx}$, and solve.

By following these steps, you can confidently find $\frac{dy}{dx}$ for any implicit equation. So, keep practicing, and you'll master this skill in no time. Congratulations, you've now successfully navigated an implicit differentiation problem! Keep up the great work, and don't be afraid to try other similar problems. Math might seem hard at first, but with practice, it becomes more manageable and even fun. The key is to break down problems into smaller, more manageable parts. And remember, understanding the why behind the steps is just as important as knowing the how. Now go forth and conquer more calculus challenges!