 In this video, I want to talk about the so-called quotient rule for derivatives because the derivative of the quotient is a little bit more complicated than that of a product. So the formula is going to look like the following. If f and g are differentiable functions and g of x does not go to zero at the point we're interested in, because division by zero of course would throw everything out the window there. And with those assumptions, the derivative of f divided by g. So f divided by g prime is going to look like g of x times f prime of x minus f of x times g prime of x divided by g of x squared. So we're going to take the bottom function and we square it here in the denominator. There are some important things to pay attention to here. The original function was a quotient of two other functions. The derivative will likewise be a quotient as well. The function that was on the bottom is going to be in the bottom squared. We're not taking the derivative there. It's just g of x squared. So as you square the original one, that usually is pretty easy to remember. The part that you have to watch out for when it comes to the quotient rule is this minus sign right here. There's a subtraction going on. And when you look at the numerator, it's like I got g of x times f prime of x times f of x times g prime of x. And so when you look at that expression, the gf prime minus fg prime, that looks a lot like the product rule we saw before. It's like, okay, I got it. You got one factor times the derivative of the other and then you have that other one times the derivative of the other. That's similar, but unlike the product rule because sums and products are commutative, you can mix things up and it doesn't really matter too much. But the quotient rule is going to be a big deal. If you move this minus sign, that'll give you the wrong derivative. So who do we take the derivative of first and who do you take the derivative of second? We have to pay attention to that. Well, one way of remembering it is whoever was on top, the numerator, you're going to take the derivative of it first. And then who's on the denominator, you take the derivative of that. So there is sort of like an order to it. Like the top has a higher priority, so you take its derivative first. That's one way of remembering it. Another way is to think of the following poem because honestly for me, I mix these things up all the time. Even after keeping calculus for many, many years, I still remember the nice little poem I learned when I took my first calculus course. So if we write things a little bit differently, right? So we're going to write this as g of x and then instead of saying f prime, I'm going to say dx of f of x there. So d stands for derivative here. So we take g of x df and we're going to take fdg. That's going to do where the denominator stays the same. Why make that modification? Because then with that modification, you get the following little couplet. It's a poem for which we're going to say low d high minus high d low square the bottom and here we go. So you can leave once you have that correctly. So what does that mean? Low d high. So the function on the bottom is clearly the lower one. The one on the top is the higher one. So we read that as low d high. So you're going to take the denominator function times the derivative of the numerator. So low d high minus that's fairly straightforward. And then you're going to end up with high d low. So you take the numerator function times the derivative of the denominator function. So that gets you the correct order you want. Low d high minus high d low and then square the bottom. Here we go. You need to square the bottom. Why do we say here we go? Well, it rhymes with low. So we make sure we get things in the correct order. So say one more time low d high minus high d low square the bottom. Here we go. If you remember this poem, then you'll remember the quotient. Well, that'll be very helpful to you in the future when you have to do this. Now, if you're like me, you can put this. You can put this poem to a song to music if you prefer me personally. I think like the the square kind of makes me think of a square dance. So it's kind of like low d high minus high d low square at the bottom. Here we go. You can of course choose whatever tune is appropriate for you to remember it. And so this is a very common poem to help people remember the quotient rule. Less common is the second verse to this to this song here, which goes something like the following high d low minus low d high square the bottom. Now you die. So you want to remember that one because the thing is, if you mix up the first line that makes this messes things up. That's the part that we have to look out for. So you have to make sure that low rhymes with go. On the other hand, high rhymes with dies. So if you end with low d high, then you're going to die. That's the long the wrong quotient rule there. So unlike the product rule, I'm not going to actually prove the quotient rule at this moment. It turns out that when we equip the product rule with the chain rule, we actually can prove the quotient rule quite easily. And so to avoid a cumbersome proof, we're going to do that later. So right now what I want to do is actually apply the quotient rule to compute the derivative of the rational function 2x minus one over 4x plus three. So the high function, of course, is going to be 2x plus minus one. The low function is going to be 4x plus three. So if we want to compute the derivative of f of x, think of the poem here. We're going to get low d high minus high d low square the bottom. Here we go. We always have to do the second half of the poem so we know that we are doing the right one. So it needs to rhyme correctly without death, of course. So now we've put everything together. Now we're going to try to compute some derivatives for which the derivatives are going to be simple enough with these ones. The derivative of 2x minus one by rules we've already seen would be two. The derivative of 4x plus three is just going to be four. The derivative of linear function is just a slope. This then sits above the 4x plus three squared. So the mistake that students make the most often when they use the cultural, they mix up the cultural, the negative sign gets in the wrong spot. If you remember the poem, that solves all of your problems. The other thing I have to caution you about is when you look at something like this, temptation comes upon many of students at this moment where they're like, ooh, I have a 4x plus three. Ooh, I have a 4x plus three. I can cancel them out and I've simplified it. No, no, no, no, no. That's not right at all. Notice if I did something like the following. If I take two plus three over two, the temptation is like, I'm going to cancel out the two. I'm going to cancel out the two. This is equal to three or one plus three, which equals four or something like that. Neither of those is correct. Notice, of course, two plus three is five over two is it would be five halves or 2.5. Notice 2.5 is neither three nor four. You can't cancel terms across a fraction bar. You can only cancel common divisors. So if you had something like three times three plus two all over three, then by all means cancel the three. That's okay. The advisor can cancel on top and bottom, but not just a common term. You can't just combine like terms like you came with addition. So the temptation to cancel the 4x plus three right here is going to be very tempting. Do not do it. That's not an acceptable cancellation because after all, if we could, if we could cancel something like that, think of the quotient rule formula f of x times, excuse me, g of x times f of x prime minus f of x times g prime of x all over g of x. If you could always unilaterally just cancel out that g right there. Why would we not just do it right here in the formula? Have a simpler formula. Why would we use a formula that always simplifies up because it doesn't. So watch out for those things. Okay. That's that's that's a very common mistake with these type of things. So what we're going to do instead, you know, we're going to do like like what I like to call the right way is we're going to distribute this to we're going to distribute this for so that we can combine like terms in the numerator. We're going to get an 8x plus six. Then we're going to get a minus here an 8x minus four. Do make sure that minus sign distributes on to all of the pieces in the second part, right? This negative sign is inclusive here. So you're going to see that there's an 8x minus an 8x. They're going to cancel each other out and that'll leave behind a six from the first one and then you're going to get a negative negative four. So that's actually a positive four that combines to give us a 10 over 4x plus three squared, which is then the derivative of this function using the quotient rule that we just learned about. With one last example, let's find the equation of the tangent line to the curve given as y equals e to the x over one plus x squared. And we want to find this tangent line at the point one comma e over two. You'll notice that if we take the x quarter to equal one, we're going to end up with e to the first over one plus one squared. One plus one squared, of course, is two, so we end up with e over two. So that's the point that they're given right there. Now recall that to find the equation of the tangent line, we have to use the formula y minus f of a is equal to f prime at a times x minus a. We're in this case, our function is f of x right here, right? So the a coordinate that value is given us a here is going to equal one. So we can plug that into the formula and go with it there. We're going to get an x minus one like so. And f of one, which is what we would be computing over here is already been computed for us as well. That's going to be this e over two that we need to compute. So what's left for us mysterious still is the f prime at one. So we need to compute what that is. That's going to be our goal at the moment. So let's compute the derivative of the function f using the quotient because it is a quotient of functions going on here, right? We have this f of x equals e to the x over one plus x squared. The quotient rule is necessary for the derivative here. So we'll compute it using our poem here. So we get low d high minus high d low square at the bottom. Here we go. And so we're going to take one plus x squared and we're going to square the entire thing, which really we're going to leave it factored. So it won't be too much of a consequence for us here. So recall that when you take the derivative of e to the x, it's equal to its own derivative. So the derivative will be itself e to the x. So we get one plus x squared times e to the x. Then we're going to subtract from that e to the x times when we take the derivative of one plus x squared, the derivative of a constant is going to be zero. And the derivative of x squared is going to be a two x like so. This all sits above the one plus x squared quantity squared right there. For which if we wanted to, we could simplify this a little bit more. Notice that there's this common factor of e to the x. I'm going to factor that thing out of this expression. And so we end up with e to the x multiplied by one minus two x plus x squared. This sits above one plus x squared quantity squared. And it turns out one minus two x plus x squared that actually does factor. That's going to factor as a one minus x quantity squared. So e to the x times one minus x squared all over one plus x squared quantity squared. So we can't really, we can't really simplify the fraction in more than it is, but this would give us the correct derivative. Now our goal is not to find the derivative per se. We're trying to find the tangent line, which we know a specific slope. So we found the derivative dy over dx, but we need to evaluate it at the value x equals one. Or in other words, we're looking for f prime at one like we said a moment ago. In which case if we plug in x equals one into our function here, we'd end up with e to the first times one minus one squared all over one plus one squared. For which we simplify this, we end up with e times zero over, we're going to get a two squared, which this is all just going to equal zero. This tells us that the slope of the tangent line here is zero. It turns out that x equals one, we have now obtained a horizontal tangent line. So let's erase this and insert our slope of zero right there. Well, anything times zero is going to give us zero. So we have a y minus e over two is equal to zero. To solve for y to put it in slope intercept form, we're going to add e halves to both sides of the equation. And we end up with the equation of the tangent line will just be y equals e over two. And again, we see that this is a horizontal tangent line because it has the form y equals some constant.