字幕表 動画を再生する 英語字幕をプリント - [Voiceover] What we want to do is find the derivative of this G of X and at first it can look intimidating. We have a sine of X here. We have a cosine of X. We have this crazy expression here with a pi over cube root of X we're squaring the whole thing and at first it might seem intimidating. But as we'll see in this video, we can actually do this with the tools already in our tool kit. Using our existing derivative properties using what we know about the power rule which tells us the derivative with respect to X. Of X to the N is equal to N times X to the N minus one, we've see that multiple times. We also need to use the fact that the derivative of cosine of X is equal to negative the sine of X. And the other way around the derivative with respect to X of sine of X is equal to positive cosine of X. So using just that we can actually evaluate this. Or evaluate G prime of X. So, pause the video and see if you can do it. So probably the most intimidating part of this because we know the derivative's a sine of X and cosine of X is this expression here. And we can just rewrite this or simplify it a little bit so it takes a form that you might be a little bit more familiar with. So, so let me just do this on the side here. So, pi pi over the cube root of X squared. Well that's the same thing. This is equal to pi squared over the cube root of X squared. This is just exponent properties that we're dealing with. And so this is the same thing. We're gonna take X to the one third power and then raise that to the second power. So this is equal to pi squared over. Let me write it this way, I'm not gonna skip any steps because this is a good review of exponent property. X to the one third squared. Which is the same thing as pi squared over X to the two thirds power. Which is the same thing as pi squared times X to the negative two thirds power. So when you write it like this it starts to get into a formula, you're like, oh, I can see how the power rule could apply there. So this thing is just pi squared times X to the negative two thirds power. So actually let me delete this. So, this thing can be rewritten. This thing can be rewritten as pi squared times X to the negative to the negative two thirds power. So now let's take the derivative of each of these pieces of this expression. So, we're gonna take we want to evaluate what the G prime of X is. So G prime of X is going to be equal to. You could view it as a derivative with respect to X of seven sine of X. So we can take do the derivative operator on both sides here just to make it clear what we're doing. So we're gonna apply it there. We're gonna apply it there. And we're going to apply it there. So this derivative this is the same thing as this is going to be seven times the derivative of sine of X. So this is just gonna be seven times cosine of X. This one, over here, this is gonna be three, or we're subtracting, so it's gonna be this subtract this minus. We can bring the constant out that we're multiplying the expression by. And the derivative of cosine of X so it's minus three times the derivative of cosine of X is negative sine of X. Negative sine of X. And then finally here in the yellow we just apply the power rule. So, we have the negative two thirds, actually, let's not forget this minus sign I'm gonna write it out here. So you have the negative two thirds. You multiply the exponent times the coefficient. It might look confusing, pi squared, but that's just a number. So it's gonna be negative and then you have negative two thirds times pi squared. Times pi squared. Times X to the negative two thirds minus one power. Negative two thirds minus one power. So what is this going to be? So we get G prime of X is equal to is equal to seven cosine of X. And let's see, we have a negative three times a negative sine of X. So that's a positive three sine of X. And then we have, we're subtracting but then this is going to be a negative, so that's going to be a positive. So we can say plus two pi squared over three. Two pi squared over three. That's that part there. Times X to the. So negative two thirds minus one, we could say negative one and two thirds, or we could say negative five thirds by power. Negative five thirds power. And there you have it. We were able to tackle this thing that looked a little bit hairy but all we had to use was the power rule and what we knew to be the derivatives of sine and cosine.
B1 中級 米 Worked example: Derivatives of sin(x) and cos(x) | Derivative rules | AP Calculus AB | Khan Academy 3 1 yukang920108 に公開 2022 年 07 月 12 日 シェア シェア 保存 報告 動画の中の単語