Implicit Differentiation: Solving For Dy/dx At A Specific Point

Hey math enthusiasts! Today, we're diving deep into the fascinating world of implicit differentiation. We'll be using this powerful technique to find dy/dx (the derivative of y with respect to x) and then evaluate it at a specific point. Get ready to flex those calculus muscles! This is a core concept, and once you grasp it, you'll be able to tackle a wide variety of problems with ease. Trust me, it's not as scary as it sounds. We'll break it down step-by-step, making sure you understand every single move.

Implicit differentiation is your go-to method when dealing with equations where y isn't explicitly defined as a function of x. Think of it like this: Sometimes, you can't just isolate 'y' on one side of the equation. That's where implicit differentiation saves the day! Instead of trying to rearrange the equation, we directly differentiate both sides with respect to x, treating 'y' as a function of x. This approach is super useful when dealing with equations like circles, ellipses, or anything where 'y' is mixed up with 'x' in a more complex way. It's an essential skill for anyone studying calculus because it unlocks the ability to solve a vast array of problems that would be impossible to solve with explicit differentiation alone. We will also evaluate the derivative at the specified point, which gives us the slope of the tangent line at that specific location on the curve represented by the equation. So, not only will we find the derivative, but we will also understand what the derivative represents graphically: how the function is changing at a specific point.

The Mechanics of Implicit Differentiation

Let's get down to the nitty-gritty. The core idea is simple: differentiate both sides of the equation with respect to x. When you encounter a term involving 'y', remember to apply the chain rule. This means multiplying by dy/dx after differentiating the 'y' term. The chain rule is the secret ingredient here. For instance, if you have y², its derivative with respect to x is 2y * dy/dx. This dy/dx is crucial; it tells us how 'y' is changing with respect to 'x'. It's the key to understanding the relationship between the two variables when they're intertwined in an implicit equation. Remember that the goal is to get dy/dx by itself. This usually involves some algebraic manipulation—collecting all the dy/dx terms on one side and factoring it out. Once you've done that, you can solve for dy/dx in terms of x and y. This process is like a puzzle: each step brings you closer to the final solution. The more you practice, the faster and more comfortable you'll become. Let's start with our equation xy = 35. We need to find dy/dx by implicit differentiation. We will also evaluate this derivative at the point (-5, -7).

Step-by-Step: Finding dy/dx for xy = 35

Alright, let's roll up our sleeves and solve this problem. We'll break down the process into manageable steps. This will make it easier to follow and understand. Remember, the key is to stay organized and pay attention to each step. We'll be using the product rule, so let's recall it. The product rule states that the derivative of a 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. If we have two functions, u(x) and v(x), then the derivative of u(x)v(x) is u'(x)v(x) + u(x)v'(x).

1. Differentiate Both Sides with Respect to x

First, we differentiate both sides of the equation xy = 35 with respect to x. On the left side, we need to apply the product rule since we have a product of two functions of x, which are x and y. The product rule states that the derivative of a 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. So, the derivative of xy with respect to x is (1 * y) + (x * dy/dx), and the derivative of 35 is 0 because 35 is a constant.

Therefore, we have:

• d/dx (xy) = d/dx (35)
• y + x * dy/dx = 0

Remember, we treat y as a function of x, so when we differentiate y with respect to x, we get dy/dx. This is a crucial step! It’s where the implicit differentiation magic happens. Now we are on the right track; the next steps will be easy.

2. Isolate dy/dx

Now, our goal is to isolate dy/dx. To do this, subtract y from both sides of the equation y + x * dy/dx = 0. This gives us:

• x * dy/dx = -y

Next, divide both sides by x to solve for dy/dx. This yields:

• dy/dx = -y/x

Boom! We've found the derivative. Now we have dy/dx in terms of x and y. We're well on our way to solving the problem, and we're now one step away from finding the slope of the tangent at a specific point. Remember to double-check each step to make sure you haven’t made any mistakes. The key is to be methodical and careful throughout the process.

3. Evaluate the Derivative at the Point (-5, -7)

We have successfully calculated dy/dx = -y/x. Now we are going to evaluate the derivative at the point (-5, -7). To do this, substitute x = -5 and y = -7 into our derivative expression. This means replacing every 'x' with -5 and every 'y' with -7.

• dy/dx = -(-7) / -5
• dy/dx = 7 / -5
• dy/dx = -7/5

So, the value of the derivative dy/dx at the point (-5, -7) is -7/5. This is the slope of the tangent line to the curve xy = 35 at the point (-5, -7). It tells us how the function is changing at that specific location. We are done! Great job! We've tackled the problem and gained a deeper understanding of implicit differentiation. By working through these steps, we've demonstrated how to find the derivative using implicit differentiation and how to evaluate it at a given point.

Graphical Interpretation

The result we just found, -7/5, represents the slope of the tangent line to the hyperbola xy = 35 at the point (-5, -7). The hyperbola is a curved shape, and the tangent line touches the curve at this specific point without crossing it. The slope tells us the rate of change of the function at that point. Specifically, a slope of -7/5 means that for every 5 units we move to the right along the x-axis, we move down 7 units along the y-axis (or vice versa, move 5 units left and 7 units up). Graphically, you could visualize the tangent line touching the curve at the point (-5, -7) and sloping downwards with this rate of change. This gives us a local understanding of how the curve behaves near that specific point. Imagine zooming in on the point (-5, -7) on the curve; the tangent line is the closest linear approximation of the curve at that point.

Conclusion: Mastering Implicit Differentiation

Awesome work, guys! We've successfully navigated the world of implicit differentiation and evaluated the derivative at a specific point. You've now got a solid understanding of how to find dy/dx when dealing with implicit equations. Remember, practice is key! The more you work through different examples, the more comfortable and confident you'll become. Keep exploring, keep questioning, and keep having fun with math! You now have a valuable tool in your calculus toolkit. This is going to be useful for a variety of problems, so it's a worthwhile skill to hone. So, go forth and conquer those implicit differentiation problems! And remember, if you get stuck, don't worry. Review the steps, try another example, and don't be afraid to ask for help.

Here’s a quick recap of the key steps:

• Differentiate both sides: With respect to x, remembering the chain rule for terms involving y.
• Isolate dy/dx: Using algebraic manipulations.
• Evaluate at the point: Substitute the given x and y values into the derivative.

Keep practicing, and you'll be acing these problems in no time! Keep up the great work, and don't hesitate to ask any questions. Happy calculating! You've got this!