 Let us find together the derivative of y with respect to x, where x and y are given by the equation relationship you see here on the screen, sine of x plus y is equal to y squared times cosine of x. Now, before your trig identity sense goes crazy right here, we're not gonna need to use the sum of angles identity for sine because we don't need to solve for y in order to calculate the derivative. We're gonna calculate the derivative implicitly. Now, to calculate the derivative implicitly, what we're essentially gonna do is we're gonna take the derivative of the left-hand side with respect to x, and we also take the derivative of the right-hand side with respect to x. The idea here is if you have an equation, as long as you do the same thing to the left-hand side as you do to the right-hand side, what's good for the goose is good for the gander, then equality will be maintained. So if we take the derivative of the left and take the derivative of the right, then we can find the derivative. We can solve for dy over dx in this situation. Now, for the sake of simplicity, you're gonna see me using this little prime notation. When you see that prime notation, notice this is just an abbreviation for taking the derivative with respect to x. So for example, we need to take the derivative of sine of x plus y prime, that is, we're taking the derivative with respect to x. We also need to take the derivative of y squared times cosine of x with respect to x right there. Now, to take the left-hand side first, we have to find the derivative of sine of x plus y. This is where the chain rule is gonna come into play here. The chain rule is critical whenever you do any calculation with implicit differentiation. So we have basically two functions in play here. We have the sine function, which is this outer function, and sitting inside of it is this function x plus y, this function of two variables, x plus y. And when drawing these little links of the chain here on the screen to emphasize why this thing is called the chain rule, we have these two functions we have to take care of. So we first take the outer derivative. What's the derivative of sine? And as many of us know, the derivative of sine is cosine. The angle when you take the derivative here will stay the same. It's the original angle of x plus y. But then when you take the inner derivative, that is the derivative of the inside function, we have to take the derivative of x plus y right here. And so I don't wanna do too much all at once, so I'm just gonna leave it there, x plus y prime. Now, on the right-hand side, this is the advantage by doing the derivative on both sides. It can actually work on them kind of simultaneously because the calculation of one doesn't exactly affect the calculation of the other. So looking on the right-hand side, what derivative rule would come into play for y squared times cosine of x? Well, the answer's in the question there. It's y squared times cosine of x. We have these two factors in play. The product rule will come into play. So the derivative of the right-hand side will look like y squared prime times cosine of x. And then you add to that y squared times cosine of x prime. Remember, in all of these calculations, we're taking the derivative with respect to x. On the left-hand side, we have this factor of cosine of x plus y. We don't need to do anything with it yet. But we still have the x plus y prime. By derivative rules, if you take the derivative of sum, you can take the derivative of e to the sum and separately. So we're gonna get x prime plus y prime in that calculation. On the right-hand side, we have some derivatives to calculate. What do you do with a y squared prime? Well, this is the key of implicit differentiation. That is to say, implicit differentiation is based upon the chain rule. If you're trying to take the derivative of y squared with respect to x, you might be like, I don't know how y is related to x. So I'm kind of trapped. But the chain rule is like, no, I actually can give you an escape. It seems kind of ironic, you know, the chains are usually used for confining people. But in calculus, the chain rule is actually quite liberating. When you take the derivative here, you're gonna take the derivative with respect to y, and then you take the derivative of y with respect to x with the understanding that the dy's would cancel out, given us the original expression. When you take the derivative of y squared with respect to y, by the power rule, you get a two y. But then you need that inner derivative, right? You get this dy over dx, which we just call y prime for short, times that by cosine of x. And then with the second term, the y squared here times the cosine of x prime, well, you're taking the derivative of cosine of x with respect to x. This is the calculations we've seen many times before. The derivative would be a negative sine of x. Don't forget the negative sign right there. And so the last things to consider here is that notice x prime, what's the prime notation mean again? It just means take the derivative with respect to x, dx over dx is a one. So we could write that as a one. The y primes, we just will leave them alone for now. And so in complete simplified form, the left hand side becomes cosine of x plus y times one plus y prime. The right hand side will look like two y, y prime, cosine of x plus, well, I should say minus, right? Bring out the negative sign there, y squared sine of x. And so now we've calculated all of these derivatives, right? So there's no more calculus here. That the rest of this problem is purely just going to be algebra. We need to solve for the y prime. Now, the nice thing here is that when you're calculating these derivatives implicitly, you will always get that with respect to y prime, it's a linear equation. That is the only y primes that show up are the first power and that's it. So we wanna solve for y prime. In order to do that, we should move all of the y primes to one side of the equation by addition or subtraction. So I'm gonna move this one to the left hand side by subtracting both sides, the expression two y, y prime, cosine of x. This is gonna give us cosine of x plus y times one plus y prime. Actually, now I think of it, I need to distribute that cosine here so I can free up the y prime. So let's do that, distribute it onto the one. There's gonna be a cosine of x plus y, but you'll also get a cosine of x plus y times y prime. Now we have a minus two y, y prime, cosine of x. And then on the right hand side, we have this negative y squared sine of x. So our goal was to get everyone who's a multiple of y prime on the left hand side and then exclude everyone else. So this cosine of x plus y needs to move to the right hand side. We'll do that by subtracting cosine of x plus y from both sides. So now on the left hand side, everyone should be a strict multiple of y prime. We have the cosine of x plus y times y prime. We have the negative two y, y prime, cosine of x. And then on the right hand side, we have this negative cosine of x plus y. We have this minus y squared sine of x. So we've gathered everyone on the right hand side who's not a multiple of y squared, y prime, I should say, excuse me. And so since everyone is a multiple of y prime on the left hand side, we can factor it out. And upon doing so, we'll get a y prime times this expression, this trigonometric expression, cosine of x plus y minus two y cosine of x. The right hand side just stays what it was a moment ago. I'm just gonna factor out the negative sign. Cosine of x plus y plus y squared sine of x. Like so. And so this expression, the cosine of x plus y minus two y cosine of x. You can think of this as the coefficient of the y prime. So if we divide both sides by said coefficient, we'll do exactly that. And that'll then leave y prime all by itself on the other side of the equation. Like so. So the left hand side now becomes y prime, which remember y prime is just an abbreviation for dy over dx, the derivative of y with respect to x. And so on the right hand side, we're gonna have this whole thing. Now to kind of avoid this extra minus sign, I'm actually gonna distribute the minus sign across the denominator here. Negative one's this curious number that multiplying by negative one is actually the same thing as dividing by negative one. So that's why you can kind of distribute the negative sign across the numerator or denominator. Just so I just clean it up a little bit. And so then the numerator will look like cosine of x plus y plus y squared sine of x. And then the denominator will look like to y cosine of x minus cosine of x plus y. And so now you can see that even with this, this implicit trigonometric equation between x and y, we were able to still calculate the derivative of the variable y with respect to x. And so if you've been following along with our lecture series, this does bring us to the end of lecture 24. The good news is in lecture 25, we're gonna continue our discussion of implicit differentiation with some more examples. You know, more challenging even than the one we see in this video right now. So take a look for that link if you wanna see some more examples. If you've learned anything from this video or other videos in this series, feel free to hit the like button and also feel free to subscribe if you wanna see more videos like this in the future. As always, if you have any questions whatsoever about things you see in these videos, feel free to post your questions in the comments and I'll gladly answer them at my soonest convenience. Bye everyone, have a great day.