12/14/2023 0 Comments Product rule calculus khan academy![]() ![]() One times g of negative one, g of negative one minus f of negative one times g prime of negative one. It's going to be f prime of negative one, lowercase f prime, that'sĪ little confusing, lowercase f prime of negative Negative one is equal to, well everywhere we see an x, Let's just say we want to evaluate F prime when x is equal to negative one. To evaluate this thing, and you might say, wait, howĭo I evaluate this thing? Well, let's just try it. Prime of x, and likewise, you could say, well that is You could say this is the same thing as g Of writing this with a derivative operator, In the denominator, so times g of x, minus theįunction in the numerator, minus f of x, not taking its derivative, times the derivative in theįunction of the denominator, d, dx, g of x, all of that over, so all of this is going to be over the function in the denominator squared. Of the quotient rule, its derivative is going toīe the derivative of the function of the numerator, so d, dx, f of x, times the function In the denominator, we can say its derivative, and this is really just a restatement So if you have some function defined as some function in the numerator divided by some function Gonna state it right now, it could be useful to know it, but in case you ever forget it, you can derive it pretty quickly from the product rule, and if you know it, the chain rule combined, you can get the quotient rule pretty quick. So if we want to take it's derivative, you might say, well, maybe the quotient rule is important here. And the way they've set up capital F, this function definition, we can see that it is a quotient of two functions. So the way that we can do that is, let's just take theĭerivative of capital F, and then evaluate it at x equals one. Of x, and they want us to evaluate the derivative of capital F at x equals negative one. Let capital F be a function defined as, so capital F is definedĪs lowercase f of x divided by lowercase g ![]() Let g be the function g of x is equal to two x to the third power. So hopefully this makes the product rule a little bit more tangible.A function such that f of negative one is equal to three, f prime of negative one is equal to five. 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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