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  • - [Voiceover] Let's say that

  • y is equal to five minus three x over x squared plus three x

  • And we want to figure out

  • what's the derivative of y in respect to x.

  • Now it might immediately jump out at you that look,

  • y is being defined as a rational expression here

  • as the quotient of two different expressions.

  • We could even view this as two different functions.

  • You could view this one up here as u of x,

  • so you could say this is the same thing,

  • this is the same thing as u of x, over,

  • you could view the one in the denominator as v of x.

  • That one right there is v of x

  • and so if you're taking the derivative of something

  • that can be expressed in this way,

  • as the quotient of two different functions,

  • well then you could use the quotient rule.

  • I'll give you my little aside, like I always do

  • the quotient rule, if you ever forget it

  • it can be derived from the product rule

  • and we have videos there,

  • cause the product rule's a bit easier to remember.

  • But what I can do is just say,

  • "look, dy dx, if y is just u of x over v of x",

  • I'm just gonna restate the quotient rule.

  • This is going to be, this is going to be

  • the derivative of the function in the numerator, so.

  • d, dx of u of x

  • times the function in the denominator

  • times v of x minus, I'll do the

  • minus the function in the numerator, u of x,

  • times the derivative of the function in the denominator

  • times d dx, v of x

  • and we're almost there

  • and then over, over the function in the denominator squared

  • the function in the denominator squared.

  • So this might look messy but all we have to do now

  • is think about what is the derivative of u of x?

  • What is the derivative of v of x?

  • And we should just be able to substitute those things

  • back into this expression we just wrote down.

  • So let's do that.

  • So the derivative with respect to x of u of x, of u of x

  • is equal to, let's see, five minus three x.

  • The derivative of five is zero.

  • The derivative of negative three x,

  • well that's just gonna be negative three.

  • That's just negative three.

  • If any of that look completely unfamiliar to you

  • I encourage you to review the derivative properties

  • and maybe the power rule.

  • Now let's think about what is the derivative

  • with respect to x, derivative with respect to x,

  • of v of x, of v of x?

  • Well derivative of x squared,

  • we just bring that exponent out front,

  • it's gonna be two times x to the two minus one

  • or two x to the first power or just two x

  • and then the derivative of three x is just three.

  • So two x plus three.

  • Now we know everything we need to substitute back in here.

  • The derivative of u with respect to x,

  • this right over here is just negative three.

  • V of x, this we know is x squared plus three x.

  • We know that this right over here is v of x.

  • And then u of x we know is five minus three x.

  • Five minus three x.

  • The derivative of v with respect to x

  • we know is two x plus three, two x plus three.

  • And then finally v of x, we know is x squared plus three x.

  • So this is x squared plus three x

  • and so what do we get?

  • Well we are going to get, it's gonna look a little bit hairy

  • It's going to be equal to negative

  • I'll focus this so first we have this business up here.

  • Negative three times x squared plus three x.

  • So I'm just gonna distribute the negative three.

  • So it's negative three x squared minus nine x

  • and then from that we are going to subtract

  • the product of these two expressions and so let's see,

  • what is that going to be?

  • Well, we have a five times two x,

  • which is ten x

  • a five times three which is 15

  • we have a negative three x times two x

  • so that is going to be negative six x squared

  • minus six x squared, and then a negative three x times three

  • so negative nine x.

  • And let's see we can simplify that a little bit.

  • Ten x minus nine x,

  • well that's just going to leave us with x.

  • So ten x minus nine x is just going to be x

  • and then in our denominator, we're almost there.

  • In our denominator we could just write that as

  • x plus three x, x squared plus three x squared

  • or if we want we could expand it out,

  • I'll just leave it like that

  • x squared plus three x squared

  • and so if we wanna simplify

  • or attempt to simplify this a little bit.

  • It's going to be negative three x squared minus nine x

  • and then you're gonna have a negative minus x

  • minus x and then minus 15

  • and then minus negative six x squared

  • so plus six x squared, all of that over

  • x squared plus three x squared

  • or x squared plus three x, squared.

  • I should say it that way.

  • Now let's see this numerator I can simplify a little bit.

  • Negative three x squared plus six x squared

  • that's going to be positive three x squared

  • and then we have in orange,

  • we have negative nine x minus an x,

  • well that's gonna be minus ten x, minus ten x

  • and then we have minus 15.

  • So minus 15.

  • So there you have it, we finally have finished.

  • This is all going to be equal to

  • this is all going to be equal to

  • three x squared minus ten x minus 15

  • over x squared plus three x, squared

  • and we are done.

- [Voiceover] Let's say that

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B1 中級

Differentiating rational functions | Derivative rules | AP Calculus AB | Khan Academy

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    yukang920108 に公開 2022 年 07 月 12 日
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