 let us find the derivative of y with respect to x. If x and y are determined by the relationship, x plus the square root of x times the square root of y is equal to y squared. You'll notice that in this relationship, we don't have something of the form y equals f of x, that is y is not explicitly solved for in this relationship, which is okay, right? We can compute the derivative using implicit differentiation, which is what we're gonna demonstrate here in this video. Try to solve for y, assuming that's even algebraically possible, because sometimes you do get problems with that. It's a chore that is not really necessary. We can actually derive dy over dx completely implicitly here. So what we're gonna do is we're gonna take the derivative with respect to x on both sides of the equations. We're gonna take the derivative of the left-hand side, and we're gonna do this on the right-hand side as well, for which on the left-hand side, we have to take the derivative of x plus the square root of x times the square root of y. I'm gonna put a little prime notation here as just shorthand for the derivative with respect to x. Keep that in mind here. As we do have the two variables, it's important we remember we're taking the derivative with respect to x. And then we have to take the derivative of the right-hand side as well. So we're gonna take the derivative of y squared with respect to x. Now on the left-hand side, because we have a sum of two terms, we have an x and the square root of x times the square root of y. By derivative laws, we can take their derivative separately. So we're gonna get an x prime plus the derivative of the square root of x times the square root of y, like so. On the right-hand side, this is where the chain rule comes out, which is really the basis behind implicit differentiation. We wanna recognize that we're taking the derivative with respect to x, but we don't have an x there, we have a y. So when we take the derivative with respect to x of this y squared, we can use the chain rule and be like, oh, I could do the following. I could take the derivative of y squared with respect to y and then times that by the derivative of y with respect to x, for which if you take the derivative of y with respect, y squared with respect to y, by the power rule, you'll get a two y. And then when you take the derivative of y with respect to x, well, that's just the y prime we're searching for right now. So we end up with just this y prime on the right-hand side. Continuing on, when we look at the left-hand side again, remember when you see this x prime, the prime is just short for the derivative with respect to x. So this is just dx over dx, which is equal to one. So we can make that simplification right here. We get one plus. Well, now we have to take the derivative of the square root of x times the square root of y. Because we have a product of two functions, we should take the derivative using the product rule for which we'll get the square root of x prime times the square root of y. And then we add to that the square root of x times the square root of y prime. We have to take those derivatives in just a second. And then on the left-hand, on the right-hand side, we still have this two y, y prime, not much to do there. So now we have to take the derivative of the square root of x. So remember, when you're taking the derivative of the square root, say, of u with respect to x, the idea here is to think of it as a power function. So we have u to the one-half power. And so by the usual power rule, you're gonna get one-half u to the negative one-half power times the derivative of u. That is the derivative of u with respect to x. For which you could rewrite that, and this is how we're gonna do it here, you're gonna end up with a y, u prime over two times the square root of u. So that's the derivative, that's the derivative of square root function using the chain rule. So when we apply that here, we're gonna get one plus, the square root of x's derivative becomes, well, you get a two times square root of x in the denominator, you also have to get the derivative of x on the top. But like we observed earlier, that's just a one, right? So I can actually get away with this putting a one right there. We times that by the square root of y. And then on the x, so we have the square root of x, we have to multiply this by the derivative of the square root of y, which as we observe from this formula right here, we end up with a y prime over two times the square root of y in that fashion. This is still equal to two y, y prime. So now that we've done this, we've calculated all the derivatives. We're now want to solve for y prime, because that was our goal, remember? Looking at the top of the screen, our goal was to find dy over dx, y prime is just an abbreviation of that. So looking at the current equation, what y primes do we have? We have a y prime right here, we have a y prime right here. So we wanna combine the y primes together and then we can solve for the y prime. So what I'm gonna do is move this expression to the right hand side and we can do that by subtracting both sides of the equation there. So on the left hand side, we'll be left with a one times the square root of y over two times the square root of x. On the right hand side, we have a two y, y prime, and then we're gonna get subtraction here minus the square root of x, y prime over two times the square root of y, like so. And so now notice the right hand side, everyone has a factor of y prime. We're gonna factor it out. That was the whole point of moving it there. So that the right hand side is now gonna look like two y minus the square root of x over two times the square root of y and you multiply that by y prime. So that's the right hand side. So then to solve for y prime, we're gonna divide both sides by the coefficient of y prime, which looks like two y minus the square root of x over two times the square root of y. We have to do it to the other side as well. What's good for the goose is good for the gander. We have to treat them the same. And so then solving for y prime, we end up with y prime is equal to what looks like a mess of numbers right now. We get one plus the square root of y over two times the square root of x all over two y minus the square root of x over two times the square root of y, which technically speaking, this is the derivative we've now solved for y prime here, but like I said, it's quite messy. This isn't very practical form for the derivative. Notice we have these fractions inside of fractions. Let's try to clean that up a little bit. Let's notice what we have here. We have a two times the square root of x and we have a two times the square root of y. If we identify the least common denominator of these baby fractions there, you'll end up with a two times the square root of x times the square root of y. So what I wanna do is multiply the top of the main denominator by two root x, root y, and then we wanna do that to the denominator as well so things stay proportional. You'll notice of course that this right here in red is just the number one written strategically here to kinda help us clean up these fractions. So when we distribute this across the top and the bottom, this will clean up our fractions a lot. We're gonna end up with a two times the square root of x times the square root of y. When you multiply on the fraction next, the two and the square root of x will cancel out and so then you're gonna get a square root of y times the square root of y which that simplifies just to be a y. Like so. We're gonna see something very similar in the denominator. We're gonna get a two times two which is a four. You have a y, the square root of x times the square root of y. Just leave that alone. I mean if you wanna combine the y's together you can get y to the three halves power. I'm just gonna leave it as y times the square root of y. I think that's perfectly simple for us right here. And then when you multiply, when you distribute this onto the second fraction right here, again where the twos will cancel, the square of y's will cancel and then we'll end up with a minus the square root of x times the square root of x, which we saw above that product should just be an x. The square root should disappear. So now we've cleaned up the fractions and so we see that dy over dx is equal to this thing right here. We could continue forward if we wanted to say like rationalize a numerator and denominator but I don't really see any algebraic benefit of doing that at this moment. Nothing simplified I would say happens from rationalizing. So we're gonna leave it in this format and report this as the derivative of y with respect to x which we computed using implicit differentiation.