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  • - [Voiceover] What I hope to do in this video

  • is give you a satisfying proof of the product rule.

  • So let's just start with our definition of a derivative.

  • So if I have the function F of X,

  • and if I wanted to take the derivative of it,

  • by definition, by definition,

  • the derivative of F of X is the limit

  • as H approaches zero,

  • of F of X plus H minus F of X,

  • all of that over, all of that over H.

  • If we want to think of it visually,

  • this is the slope of the tangent line and all of that,

  • but now I want to do something

  • a little bit more interesting.

  • I want to find the derivative

  • with respect to X, not just of F of X,

  • but the product of two functions,

  • F of X times G of X.

  • And if I can come up with a simple thing for this,

  • that essentially is the product rule.

  • Well if we just apply the definition of a derivative,

  • that means I'm gonna take the limit

  • as H approaches zero,

  • and the denominator I'm gonna have at H,

  • and the denominator, I'm gonna write a big,

  • it's gonna be a big rational expression,

  • in the denominator I'm gonna have an H.

  • And then I'm gonna evaluate this thing at X plus H.

  • So that's going to be F of X plus H,

  • G of X plus H and from that I'm gonna subtract

  • this thing evaluated F of X.

  • Or, sorry, this thing evaluated X.

  • So that's gonna be F of X times G of X.

  • And I'm gonna put a big, awkward space here

  • and you're gonna see why in a second.

  • So if I just, if I evaluate this at X,

  • this is gonna be minus

  • F of X, G of X.

  • All I did so far is I just applied the definition

  • of the derivative, instead of applying it to F of X,

  • I applied it to F of X times G of X.

  • So you have F of X plus H, G of X plus H

  • minus F of X, G of X,

  • all of that over H.

  • Limit as H approaches zero.

  • Now why did I put this big, awkward space here?

  • Because just the way I've written it write now,

  • it doesn't seem easy to algebraically manipulate.

  • I don't know how to evaluate this limit,

  • there doesn't seem to be anything obvious to do.

  • And what I'm about to show you,

  • I guess you could view it as a little bit of a trick.

  • I can't claim that I would have figured it out on my own.

  • Maybe eventually if I were spending hours on it.

  • I'm assuming somebody was fumbling with it long enough

  • that said, "Oh wait, wait.

  • "Look, if I just add and subtract at the same term here,

  • "I can begin to algebraically manipulate it

  • "and get it to what we all know

  • "as the classic product rule."

  • So what do I add and subtract here?

  • Well let me give you a clue.

  • So if we have plus,

  • actually, let me change this,

  • minus F of X plus H, G of X,

  • I can't just subtract, if I subtract it

  • I've got to add it too, so I don't change the value

  • of this expression.

  • So plus F of X plus H, G of X.

  • Now I haven't changed the value,

  • I just added and subtracted the same thing,

  • but now this thing can be manipulated

  • in interesting algebraic ways to get us to

  • what we all love about the product rule.

  • And at any point you get inspired,

  • I encourage you to pause this video.

  • Well to keep going, let's just keep

  • exploring this expression.

  • So all of this is going to be equal to,

  • it's all going to be equal to the limit

  • as H approaches zero.

  • So the first thing I'm gonna do is

  • I'm gonna look at, I'm gonna look at this part,

  • this part of the expression.

  • And in particular, let's see, I am going to

  • factor out an F of X plus H.

  • So if you factor out an F of X plus H,

  • this part right over here is going to be

  • F of X plus H,

  • F of X plus H,

  • times

  • you're going to be left with G of X plus H.

  • G of, that's a slightly different shade of green,

  • G of X plus H, that's that there,

  • minus G of X,

  • minus G of X,

  • oops, I forgot the parentheses.

  • Oops, it's a different color.

  • I got a new software program and it's making it hard

  • for me to change colors.

  • My apologies, this is not a straightforward proof

  • and the least I could do is change colors more smoothly.

  • Alright, (laughing) G of X plus H

  • minus G of X, that's that one right over there,

  • and then all of that over this H.

  • All of that over H.

  • So that's this part here

  • and then this part over here

  • this part over here, and actually it's still over H,

  • so let me actually circle it like this.

  • So this part over here

  • I can write as.

  • So then we're going to have plus...

  • actually here let me,

  • let me factor out a G of X here.

  • So plus G of X

  • plus G of X times this F of X plus H.

  • Times F of X plus H

  • minus this F of X.

  • Minus that F of X.

  • All of that over H.

  • All of that over H.

  • Now we know from our limit properties,

  • the limit of all of this business,

  • well that's just going to be the same thing

  • as the limit of this as H approaches zero

  • plus the limit of this as H approaches zero.

  • And then the limit of the product

  • is going to be the same thing as the product of the limits.

  • So if I used both of those limit properties,

  • I can rewrite this whole thing as the limit,

  • let me give myself some real estate,

  • the limit as H approaches zero of F of X plus H,

  • of F of X plus H times,

  • times the limit

  • as H approaches zero, of all of this business,

  • G of X plus H minus G of X,

  • minus G of X,

  • all of that over H,

  • I think you might see where this is going.

  • Very exciting.

  • Plus,

  • plus the limit,

  • let me write that a little bit more clearly.

  • Plus the limit as H approaches zero of G of X,

  • our nice brown colored G of X,

  • times, now that we have our product here,

  • the limit,

  • the limit

  • as H approaches zero of F of X plus H.

  • Of F of X plus H minus F of X,

  • minus F of X,

  • all of that,

  • all of that over H.

  • And let me put the parentheses where they're appropriate.

  • So that,

  • that,

  • that,

  • that.

  • And all I did here, the limit,

  • the limit of this sum,

  • that's gonna be the sum of the limits,

  • that's gonna be the limit of this

  • plus the limit of that,

  • and then the limit of the products is gonna be

  • the same thing as the product of the limits.

  • So I just used those limit properties here.

  • But now let's evaluate them.

  • What's the limit,

  • and I'll do them in different colors,

  • what's this thing right over here?

  • The limit is H approaches zero of F of X plus H.

  • Well that's just going to be

  • F of X.

  • Now, this is the exciting part,

  • what is this?

  • The limit is H approaches zero of G of X plus H

  • minus G of X over H.

  • Well that's just our,

  • that's the definition of our derivative.

  • That's the derivative of G.

  • So this is going to be,

  • this is going to be the derivative of G of X,

  • which is going to be G prime of X.

  • G prime of X.

  • So you're multiplying these two

  • and then you're going to have plus,

  • what's the limit of H approaches zero of G of X?

  • Well there's not even any H in here,

  • so this is just going to be G of X.

  • So plus G of X

  • times the limit,

  • so let's see, this one is in brown,

  • and the last one I'll do in yellow.

  • Times the limit as H approaches zero,

  • and we're getting very close,

  • the drum roll should be starting,

  • limit is H approaches zero of F of X

  • plus H minus F of X over H.

  • Well that's the definition of the derivative

  • of F of X.

  • This is F prime of X.

  • Times F prime of X.

  • So there you have it.

  • The derivative of F of X times G of X is this.

  • And if I wanted to write it a little bit more

  • condensed form, it is equal to,

  • it is equal to F of X

  • times the derivative of G with respect to X

  • times the derivative of G with respect to X

  • plus G of X,

  • plus G of X times the derivative of F with respect to X.

  • F with respect to X.

  • Or another way to think about it,

  • this is the first function times the derivative

  • of the second plus the second function

  • times the derivative of the first.

  • This is the proof, or a proof,

  • there's actually others of the product rule.

- [Voiceover] What I hope to do in this video

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Product rule proof | Taking derivatives | Differential Calculus | Khan Academy

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    yukang920108 に公開 2022 年 07 月 12 日
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