If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:7:10

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 is 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 or we have to derive this implicitly or take the derivative of it implicitly so let's apply our derivative operator to both sides so 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 3 times that's something 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 5 times the derivative of x squared is just going to be 2x times y squared 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 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 derivative of Y with respect to X so 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 that is when I applied the derivative operator similarly similar similarly that is that and when I applied the derivative operator I got that you want D the derivative with respect to X right over there so let's see if we can somehow solve for dy/dx so 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 onto this one right over here you get plus let's see 2y times 3 is 6 y times x squared x squared plus y squared let me make sure 2y times 3 6 y times x squared plus y squared squared and then I'll keep the dy/dx in that green color dy/dx is equal to is equal to well we could 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 10 X Y squared and then 5 times all of this right over here is going to be plus 10 x squared y dy DX dy dy DX and 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 10 x squared Y dy/dx from both sides n X 10 x squared y dydx 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 then I'm going to subtract this business the 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 what are we then what are we then left with well these 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 10 x 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 10x Y squared minus 6x times x squared plus y squared squared and now if we want to solve for dydx 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 do this 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 is so you could say what's the you could say maybe the X is this value right over here and then you could solve for y to figure out what this the y value would be and then you would chunk it into that x and y into this hairy expression right over here to figure out the slope to figure out the slope of the tangent line