 Hello and welcome to a screencast about doing derivatives of quotients. So the quotient rule, hopefully you guys have been practicing the product rule because this rule is very similar with two major exceptions. So the first exception is obviously we have a fraction, okay? So that right here is the major exception. The other exception with the quotient rule is the negative sign in here. So that means the order that we do things in the numerator here does matter because obviously if you've ever balanced a checkbook, if you do 2 minus 3 or 3 minus 1, that's going to make a big difference in what you're doing. So the assumptions here are saying that f of x and g of x are differentiable functions that have derivatives f prime and g prime respectively. Their quotient q of x, which is written f of x over g of x, is also a differentiable function for all x where g of x cannot be zero, because g of x is in the denominator and we all know dominators can be zero. So q prime of x, the quotient rule is g of x times f prime minus f of x times g prime of x all over g of x squared. There are a few mnemonic devices that you may have heard before. You can look up if you want to. But like I said, the order of these really does matter. So I always start with the denominator and then I know to multiply by the derivative of the numerator. Then you do the numerator by the derivative of the denominator. So just as long as you remember which order to start in, you should be fine. Okay, our first example here in the direction, say, find the derivative of each function using proper notation. Our function we're given is b of x equals 3x to the 10th minus 4x to the 6, plus 15x minus pi all over x. So hopefully you can identify that this is a quotient. I'm just going to start by typing that quotient rule here. This obviously has a numerator and a denominator. So I'm going to call this numerator f and I'm going to call this denominator g. So since we want to practice a quotient rule, we're going to go ahead and do that. But I'm also going to show you how we can throw a little bit of algebra at this one and we could have made it even easier. Okay, so it just depends on what you prefer to do. If you want to practice a quotient rule or if you want to do algebra to simplify first, by all means, go ahead. Okay, so b prime of x. And let's go ahead and identify with these derivatives r before we get too far ahead. So if this is f, then f prime is going to be, well, this is, you know, you can look at this and it's got some sums and differences in here. I see, let's see, I see a power rule and then I also see a constant back here at the end, but we'll talk about that in a sec. Okay, so derivative of 3x squared is 30x to the 9th. The derivative, and that's because if you do 10 times 3, that gives us 30 and reduce your power by 1, that gives you a 9. Minus the derivative of 4x to the 6, multiply the 4 by the 6, we get 24, reduce our power by 1, that's going to give us x to the 5th. And then plus the derivative of 15x is 15. Pi's a constant, 3.14, blah, blah, blah. So when you do the derivative of that, that then ends up giving you 0. Okay, then the derivative of g prime, so the derivative of x is just 1, so g prime is just 1. Okay, so putting those pieces together then into our formula for the quotient rule, g of x, so that means we're going to take x, multiply it by f prime. So f prime is this polynomial that we found over here, 30x to the 9th minus 24x to the 5th plus 15 minus, okay. f, that was our original function in the numerator, 3x to the 10th minus 4x to the 6 plus 15x, oopsie, x, that was my x, minus pi times the derivative of our denominator, which was 1. And all of this big, crazy messiness here is all divided by the denominator squared. Okay, now to simplify things from here, this is just some algebra, so distribute your x through, multiply your negative through, combine your like terms, and then divide each piece by x squared, okay. Or, so this would be a perfectly legitimate answer. Or if you want to do some algebra, does anybody see how we could simplify b of x first? Well, if you remember, again, back to your algebra days, we can take each piece in our numerator and since we only have one piece in our denominator, we can divide each piece out. So b of x could actually look like, okay, so again, I'm going back to the original function. We're not going to look at those derivatives anymore. 3x to the 10th over x minus 4x to the 6 over x plus 15x over x minus pi. Okay? So is this any easier? I don't know, it just depends on how much you want to memorize the quotient rule. Now we will do an example where we're forced to use it soon enough, so don't worry about that. So when we work out the algebra here, let's see, that gives us 3x to the 9th minus 4x to the 5th plus 15. And then minus, now pi over x can't really be simplified, but remember, this looks kind of like a power, so we're going to want to bring that denominator up and go ahead and make that to the negative one power. So now b prime of x is going to give us, okay, apply the power rule to each piece. And remember these constants just get multiplied. So 3 times 9 gives us 27x to the 8th minus 4 times 5 gives me a 20x to the 4th plus, well let's see what happens to that 15. 15 is a constant, so that piece drops out. So then that one goes away. So we have minus pi times negative 1, so that's a positive pi, and then x to the negative 2, because negative 1 minus 2 gives us, or sorry, negative 1 minus 1 gives us negative 2, there we go. Okay, so this would also be a perfectly legitimate answer. So I leave it to the reader or the listener here to convince me, are these two answers actually the same? And the answer is yes. It's just you took two different ways in order to figure out what this derivative is, and if that confuses you, then just pick whichever way makes more sense to you. But again, I prefer to kind of go with the second route, because to me, these derivatives are a lot easier than trying to remember what the quotient rule is. Okay, fantastic. But now I come to a function here, v of t, that's 3t minus 2t to the fourth plus the square root of t, all over t squared minus seven. So our little trick on that last example will not work with this one. You cannot break up each piece just because how the vision works, okay? So that just isn't possible. So for this one, we're gonna call the numerator f and the denominator g. So let's go ahead and figure out what the derivatives are for each of these. So f prime is, you notice I threw a, what's the first piece here? Exponential, right? So what's the derivative of 3 to the t? Well, that's 3 to the t times the natural log of 3 minus, what's the derivative of 2t to the fourth? That would be 8t to the third, because 2 times 4 gives me the 8 in front. Drop my power down by 1, that gives me a 3. Plus, okay, the square root of t, remember that's t to the 1 half power. So then the derivative of t to the 1 half is 1 half t to the negative 1 half. Cuz drop that 1 half down front, technically we multiply by 1. Then 1 half minus 1, well it's 1 half minus 2 half, gives me a negative 1 half. Okay, then g, the derivative, or the denominator here, that's t squared minus 7, that derivative is just 2t. Because we bring the 2 down front and then reduce that power by 1, that gives us a 1 technically. And then minus 7, that drops out because of the fact that it's a constant. So that one goes away. Okay, so now let's go ahead and write out that quotient rule again. So which one do we start with again? We start with the denominator. So that's g times f prime minus f times g prime all over g squared. Okay, so this is a little bit of a shortcut way to write it, similar to how I put it up above. Okay, so now let's go ahead and put those pieces together. So g of x or g of t in this case, so that's my denominator. So in this case, that's t squared minus 7 times f prime. So that's all this junk we found up here in the numerator for the derivative. So 3 to the t ln of 3 minus 8t cubed plus 1 half t to the negative 1 half minus, okay, f, so that was my original function in my numerator. 3 to the t, oopsie, minus 2t to the fourth plus the square root of t. And then times the derivative of my denominator, which was 2t. And all of this junk is divided by the original denominator, t squared minus 7 squared. And that is your answer. Now, again, depending on what your instructor expects out of you, you may have to multiply some of this stuff out, but that's algebra. I'll leave that one to the listener. Thank you for watching.