 In this video, we're gonna compute the derivative of the function x squared times e to the x all over seven x minus nine. We're gonna take the derivative with respect to x right here. So you'll notice the first thing, when you look at this function, it's a quotient function. That is we have x squared times e to the x divided by seven x minus nine. So when we calculate the derivative of this thing, we're gonna have to use the quotient rule. So remember our high function, the numerator is x squared times e to the x, our low function, the denominator is seven x minus nine. So we're going to get low d high, take the derivative of x squared times e to the x, minus high d low. So take the derivative of seven x minus nine, square the bottom, here we go. It's our little brine we say every time we do the quotient rule. And so using the quotient rule, we get that derivative. So there's some things that have to be calculated a little bit more, right? So you look at seven x minus nine, we have to take its derivative. That's not so bad, it's a polynomial function. It's just a linear function in fact, its derivative will be its slope, which is seven. On the other hand, you have to take also the derivative of x squared times e to the x, for which that derivative will unavoidably require the product rule. And so we see in this example that sometimes you have to combine various derivative rules to calculate the derivative of a function. This one we have to combine the product rule with the quotient rule here. We used the quotient rule forced. Now we're gonna use the product rule on x squared times e to the x. So we have seven x minus nine, we take the derivative of x squared e to the x. So what is that gonna look like? Well, by the product rule, we're gonna take the derivative of x squared, which of course we know to be two x. Then we multiply that by e to the x. Then we're gonna add to that x squared times the derivative of e to the x, which we know the derivative of e to the x is in fact just e to the x itself. So we can drop that down. We still have the minus x squared e to the x. And like I said a moment ago, when you take the derivative of the polynomial of the linear function, I should say, the derivative will just be at slope. So we get a seven right there. This all sits above the seven x minus nine squared. All right, don't be tempted to cancel out this seven x minus nine with any on the denominator. That's not valid because the only way we can cancel seven x minus nine from the denominator is if the entire numerator were divisible by seven x minus nine, for which this has no seven x minus nine factor to contribute. So don't fall into that temptation. Instead, I suppose we can distribute things and start combining. It is a good idea to try to factor your derivatives because that'll be very useful in future applications. So some things I can notice, there are some common divisors that could factor out. Like notice everything is divisible by e to the x. We could factor that out. We also can afford to take out an x away from everything. And so upon doing so, we factor out this, we factor out of everything, this x times e to the x, that then leaves behind, we have the seven x minus nine. We haven't multiplied that through yet. With the two x e to the x, we took away the x and the e to the x that's left behind a two. With the x squared times e to the x, we took away an x e to the x that leaves behind an x like so. And then for the other piece, again, we took away an x and the e to the x, so we have negative seven right here. This all sits above the seven x minus nine squared. So we took out the x e to the x. Those are good common factors. There's really not much else to do in the new way in terms of factorization. So we're gonna have to multiply that thing out. To do that, we're gonna take the seven x minus nine and foiler with the two plus x. So that'll give us a 14x plus seven x squared minus 18 minus nine x. There's also a negative seven x in the numerator. So let's combine like terms. You have the x e to the x already factored out. So in terms of the x squares, there's only the seven x squared. In terms of the x's, we have a 14x. Take away a seven x, that's what we seven x itself. Then take away another nine x, that should leave us with a negative two x. And then lastly, I think there's a constant of negative 18, like so. And then we have our factored function right here. Now, if we didn't factor out the x e to the x from the beginning, we could have done that later on. There's been a factor of x e to the x everywhere. It's usually better to factor things out before you expand. If there is, of course, a common denominator there, you might not see it until the end. So that's okay if you didn't see that. Now, if I were trying to simplify this a little bit more, it would be curious to know if there's a common factor on top and bottom. The denominator has a seven x minus nine squared. In the numerator, the seven x squared minus two x minus 18, is there any way of factoring that to cancel out the seven x minus nine? And with a little bit of guess and check, you'll see it's not gonna happen. If you take seven times negative 18, there's not gonna be any factors of that that add up to be negative two. Or you could also check with the discriminant for the quadratic formula. You'll see it's not a perfect square. So this thing is factor, can't simplify it anymore. So this would be probably the best form we could write the derivative in this situation where we needed both the quotient rule and the product rule.