![]() ![]() But if you don't know the chain rule yet, this is fairly useful. But you could also do the quotient rule using the product and the chain rule that you might learn in the future. Now what you'll see in the future you might already know something called the chain rule, or you might You could try to simplify it, in fact, there's not an obvious way Plus, X squared X squared times sine of X. This is going to be equal to let's see, we're gonna get two X times cosine of X. Actually, let me write it like that just to make it a little bit clearer. So that's cosine of X and I'm going to square it. All of that over all of that over the denominator function squared. The derivative of cosine of X is negative sine X. Minus the numerator function which is just X squared. V of X is just cosine of X times cosine of X. So it's gonna be two X times the denominator function. So based on that F prime of X is going to be equal to the derivative of the numerator function that's two X, right over Of X with respect to X is equal to negative sine of X. So that is U of X and U prime of X would be equal to two X. Well what could be our U of X and what could be our V of X? Well, our U of X could be our X squared. So let's say that we have F of X is equal to X squared over cosine of X. We would then divide by the denominator function squared. This exercise applies the chain rule in contexts. Get if we took the derivative this was a plus sign. The Combining the product and chain rules exercise appears under the Differential calculus Math Mission. If this was U of X times V of X then this is what we would The denominator function times V prime of X. Its going to be equal to the derivative of the numerator function. Then the quotient rule tells us that F prime of X is going to be equal to and this is going to lookĪ little bit complicated but once we apply it, you'll hopefully get a little bit more comfortable with it. So for example if I have some function F of X and it can be expressed as the quotient of two expressions. But here, we'll learn about what it is and how and where to actually apply it. It using the product rule and we'll see it has some So hopefully this makes the product rule a little bit more tangible.Going to do in this video is introduce ourselves to the quotient rule. This is the same thing as e to the x times cosine Or, if you want, you could factor out an e to the x. ![]() To the x times cosine of x, times cosine of x minus e to the x. To the x without taking it's derivative - they are That's what's excitingĪbout that expression, or that function. This right over here, you can view this as this was the derivative as e to the x which happens to be e to the x. And it might be a little bit confusing, because e to the x is its own derivative. So, times the derivative of cosine of x which is negative sine. Plus the first expression, not taking its derivative, so e to the x, times the derivative of ![]() To the x which is just, e to the x, times the second expression, not taking it's derivative, Negative sine of x, and so, what's this going to be equal to? This is going to be equal to the derivative of the first expression. ![]() And v prime of x, we know as negative sine of x. So u prime of x is stillĮqual to e to the x. One of the things that makes e so special. And if u of x is equal to e to the x, we know that the derivative of that with respect to x is still e to the x. When you just look at it like that, it seems a little bit abstract and that might even be a little bit confusing, but that's why we haveĪ tangible example here and I color-coded intentionally. Times v is u prime times v, plus u times v prime. You'll take the derivative of the other one, but not the first one. Them, but not the other one, and then the other one In each of them, you're going to take the derivative of one of So the way you remember it is, you have these two things here, you're going to end up So times v of x and then we have plus the first expression, not its derivative, just the first expression. Not the derivative of it, just the second expression. So I could write that as u prime of x times just the second expression This is going to beĮqual to the derivative of the first expression. This is going to be equal to, and I'm color-coding it so we can really keep track of things. So if we take theĭerivative with respect to x of the first expression in terms of x, so this is, we couldĬall this u of x times another expression that involves x. And let me just write down the product rule generally first. But, how do we find theĭerivative of their product? Well as you can imagine, Respect to x of cosine of x is equal to negative sine of x. We know how to find theĭerivative cosine of x. So when you look at this you might say, "well, I know how to find "the derivative with e to the x," that's infact just e to the x. And like always, pause this video and give it a go on your own before we work through it. So let's see if we can find the derivative with respect to x, with either x times the cosine of x. ![]()
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