Implicit differentiation (advanced example)

Once again, I have some crazy relationship between x and y. And just to get a sense of what this might look like, if you plot all the x's and y's that satisfy this relationship, you get this nice little clover pattern. And I plotted this off of Wolfram Alpha. But what I'm curious about in this video, as you might imagine from the title, is to figure out the rate at which y is changing with respect to x. And we're going to have to do it implicitly. We're going to have to find the implicit derivative of this. We're going to have to derive this implicitly or take the derivative of it implicitly. So let's apply our derivative operator to both sides of the derivative with respect to x on the left and the derivative with respect to x on the right. So once again, we apply our chain rule. The derivative of something to the third power with respect to that something is going to be three times that something squared. And then we have to multiply that times the derivative of the something with respect to x. So the derivative of this thing with respect to x is going to be 2x, that's the derivative of x squared with respect to x, plus the derivative of y squared with respect to y is going to be 2y times the derivative of y with respect to x. Once again, we're applying the chain rule right over here. The derivative of something squared with respect to the something, which is 2y, times the derivative of the something with respect to x, which is dy dx. Now, that is going to be equal to what we have on the right hand side. So we have a 5 times x squared times y squared. We can take the 5 out of the picture for now. Take the 5 onto the-- take it out of the derivative. The derivative of 5 times something is the same thing as 5 times the derivative. And now we can apply the product rule. So it's going to be 5 times the derivative of x squared is just going to be 2x times y squared. So that's just the derivative of the first function times the second function. Plus the first function, not taking its derivative, x squared times the derivative of the second function. Well, what's the derivative of y squared with respect to x? Well, we already figured it out. It's the derivative of y squared with respect to y, which is 2y times the derivative of y with respect to x. Let me make it clear what I just did. This is this, and then this is when I took its derivative. So that is when I applied the derivative operator. Similarly, that is that. And when I applied the derivative operator, I got that. The derivative with respect to x right over there. So let's see if we can somehow solve for dy dx. So what I'm going to do on the left hand side is I'm just going to distribute this purple thing onto both of these terms. So if you distribute this purple thing onto this term right over here, you get 3 times 2x, which is 6x, times x squared plus y squared, squared. And then if you distribute this purple stuff on to this one right over here you get plus, let's see, 2y times 3 is 6y times x squared plus y squared. Let me make sure. 2y times 3 is 6y times x squared plus y squared, squared, and then I'll keep the dy dx in that green color. dy dx is equal to, well, we can multiply the 5 times this business right over here. And so everything that's not a dy dx term maybe I will do in purple now. So you do 5 times this stuff right over here, which gives you 10xy squared. And then 5 times all of this right over here is going to be plus 10x squared y dy dx. Did I do that right? Yep, that looks just about right. And now we have to solve for dy dx. What I'm going to do is I'm going to subtract this 10x squared. I'm going to subtract the 10x squared y dy dx from both sides. 10x squared y dy dx. Derivative. That's not green. Derivative of y with respect to x. Going to subtract that from both sides so that I can get it on the left hand side. dy dx. And I'm going to subtract this business, this 6x times all this craziness from both sides. So minus 6x times x squared plus y squared, squared. Let me subtract it from here as well. Minus 6x x squared plus y squared squared. And what are we left with? Well, these guys cancel out. On the left hand side right over here, we are left with 6y times x squared plus y squared, squared minus 10x squared y times dy dx, the derivative of y with respect to x. The derivative of y with respect to x is equal to-- these characters cancel out-- and we are left with 10xy squared minus 6x times x squared plus y squared, squared. And now if we want to solve for dy dx, we just divide both sides of this equation by this business right over here. And you get the derivative of y with respect to x. And we deserve a drum roll now. The derivative of y with respect to x is equal to all of this stuff. And I'm just going to copy and paste it. So let me copy and paste it. All of that stuff over all of this stuff over all of that stuff. Once again, just going to copy and paste it. And I just have to draw the little fraction symbol. So we are done. We have figured out-- and it was kind of a hairy expression, but it didn't take us too long-- what the derivative of y with respect to x at any point is. So what you'd want to do is if you want to say, hey, what is the slope of the tangent line right at-- let me do this in a color that you can actually see. What is the slope of the tangent line right at that point? Well, you'd want to figure out what the x-coordinate of that point it so you could say maybe x is this value right over here. And then you can solve for y to figure out what the y value would be. And then you would chunk that x and y into this hairy expression right over here to figure out the slope of the tangent line.