Chain Rule Proof

by Batool Akmal

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      DLM Quotient Rule, Chain Rule and Product Rule Calculus Akmal.pdf
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    00:01 So we've spoken of two ways of dealing with the function of a function.

    00:04 We obviously have the full definition of the chain rule and also just by observation, what we can do to just differentiate faster.

    00:12 Now we said we're going to do a proof so we'll do a proof now and the proofs for chain rule are usually fairly complicated but there's a slightly simpler one that we'll talk about now which hopefully convinces you that this is true or you can just go back to our mathematical calculation and use that as a proof but either way, I'll show you both and then you can pick and choose whichever one you prefer.

    00:35 So if we move here to the chain rule proof, like I said before there are plenty of proofs out there.

    00:45 Some of them are numerically and algebraically fairly complicated but thanks to some clever people who've made things a little bit easier, I found a fairly simpler proof that we can go through that hopefully convince us of the definition.

    00:59 So, if we are to differentiate y as a function of something else, so we're wanting to differentiate this, so let's just say we want to d/dx, this function and that should then give us the gradient dy/dx. So that's what we're trying to prove here.

    01:17 Now, recall back to differentiation from first principles, where we said that dy/dx as the limit of delta x tends to zero is delta y over delta x, so that's the main definition. I'm not writing the whole function out, I'm just gonna write this for now. We now have delta y over delta x which we can re-write as the limit, as delta x tends to zero of delta y over delta u, multiplied it by delta u by delta x, so remember that's the chain rule if you're using two functions, we can re-write this in this manner.

    01:57 So we're applying that the limit of delta x tends to zero to those two functions.

    02:03 Okay now, here's a little property of limits which sets that the product of limits is the limit of the product, that means that I can break this down into two sections, so I can re-write this as the limit, delta x tends to zero of delta y by delta u and this is just the property that I'm using of limits and I can re-write this function as limit delta x tends to zero of delta u of delta x.

    02:33 So all I've done is I've split this here and this here and I applied the concept of the limit to both of them separately, they're still multiplying.

    02:41 Okay, this one is fairly simple because if you look at it, this is just delta u by delta x so change in u over change in x as delta x tends to zero.

    02:52 If you look back at differentiation from first principles, this part here will just give you, du/dx.

    03:00 So if we were to define du/dx, we'd used this definition.

    03:04 However, this little bit is slightly more problematic because we have delta y over delta u but our limit now is saying as delta x tends to zero.

    03:16 Now, this is where the proof gets a bit complicated but it's fairly easy just to change your limit to delta x to delta u.

    03:24 Okay, so all we're doing is we're making this little change here and we're changing our limit.

    03:28 So if we say, not delta x, we'll say delta u tends to zero and if we do that, you'll see that we come up with the derivative or the definition of the derivative for dy/du and again we have just decided that dy/dx is dy/du multiplied by du/dx if I just write that here what we've just derived so dy/dx is dy/du multiplied by du/dx and you will remember that this is the chain rule that we have previously derived and we have now proved using limits or the property of limits.

    About the Lecture

    The lecture Chain Rule Proof by Batool Akmal is from the course Quotient Rule, Chain Rule and Product Rule.

    Author of lecture Chain Rule Proof

     Batool Akmal

    Batool Akmal

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