 Let's find the derivative of y with respect to x. If the relationship between x and y is given by the equation, three x times y plus four y squared is equal to 10. You'll notice in this equation that we don't have a y equals some f of x right here. That is, y is not solved for explicitly here. There is a quadratic relationship here on the y's known as we have a y and a y squared. We could attempt to solve for y in order to find the derivative, but it's gonna be much easier just to compute the derivative implicitly than that's the strategy we're gonna take in this situation. So essentially, we just wanna compute the derivative of both sides of the equation. So we're gonna take the derivative with respect to x of the left-hand side, which is three x, y plus four y squared. And then we're gonna take the derivative with respect to x on the right-hand side as well of just the number 10. Now, let's actually think about the right-hand side for a moment. We're taking the derivative of 10 with respect to zero. Well, since 10 is just a constant function, 10 doesn't change as x changes. This is just gonna go over to be zero and I'm actually gonna move that to the left-hand side so, because I really don't have to do much with it anymore. As for the three x, y plus four y squared, by the usual rules of derivatives, I can take the sum of these derivatives separately. That is, we'll take the derivative of three x, y and add to that four y squared. But then also these constant coefficients can come on out of the derivative process. And so we have to take the derivative of three times x, y and then we have four times the derivative of y squared. And this is where it's very important to remember that when you use this little tick notation, this prime, this prime here is just short, just a shorthand for the derivative with respect to x. So whenever we take the derivative here, we understand we're taking the derivative with respect to x. So how do we take the derivative of the product x times y? Well, we'll use the product rule. The product rule tells us that when we take the derivative, we're gonna take the derivative of x times that by y and then we're gonna take x and times that by the derivative of y for which we have right there. So x prime y plus xy prime. When we take the derivative of the y squared, notice we're taking the derivative with respect to x, not with respect to y. y is a function of x, even if we don't have that explicit formula right now. So when we take the derivative of y squared, remember we're taking the derivative with respect to x. So therefore the chain rule comes into play. We have the outer function y squared for which when we take its derivative by the power rule, we're gonna end up with that two y. But then we also have the inner function, that is y itself, because we can think of this function, you know, being broken apart as we have this, we have this, excuse me, this x squared composed with the y right here. We put y instead of the x. So that's what we're trying to keep in mind here when we take these derivatives. So the outer derivative would be a two y and then we times that by the inner derivative, which is a y prime. And then of course the left-hand side is still equal to zero right here. So we should have a couple of y primes that showed up in this calculation. We have one right here, we have one right here. It's then gonna be our goal to try to solve for the y prime, because after all that's what we're looking for. We're looking for dy over dx. In shorthand, that's what we mean by this y prime right here. So what we're gonna do, is we're gonna move anyone who's not a multiple of y prime onto the left-hand side of the equation. Before we do that, let's distribute the three here and multiply the four by the two right here. So we end up with a three x prime y plus x, excuse me, three x y prime. And then we're gonna have an eight y y prime. Another simplification we can make right here is when you see this expression x prime right here, what is this short for? This is short for the derivative of x with respect to x, as we know that should be equal to one. So we can actually drop the x prime notation going forward here. So we're gonna subtract the negative three y from both sides of the equation. We end up with, well, negative three y on the left. On the right-hand side, we have a three x y prime plus eight y y prime. And then recognizing the derivative here as a common multiple, we can factor it out, thus making the right-hand side look like three x plus eight y multiply that by y prime. And so then to solve for y prime, we need to divide by its coefficient. That is, we need to divide both sides by three x plus eight y. Do that to the other side as well, three x plus eight y. So in the end, we get that y prime, which of course is just an abbreviation for dy over dx. This is equal to negative three y over three x plus eight y. And there's no simplification we're gonna be able to make to that fraction right there. The three on the top can't cancel with the three in the denominator because the denominator is not divisible by three. Nor can the y in the numerator cancel with the y in the denominator because the denominator does not have a common factor of y. And so we see how very nice implicit differentiation can be used to compute the derivative here, dy over dx for this function. Now, because we didn't solve for y explicitly, you will notice that the derivative involves both x and y. That's perfectly fine with implicit differentiation. We're okay having the derivative be a function of both x and y.