The previous video depending on how you're watching it, which is, if we have a function u that is continuous at a point, that, as delta x approaches zero, delta u approaches zero. But we just have to remind ourselves the results from, probably, It's written out right here, we can't quite yet call this dy/du, because this is the limitĪs delta x approaches zero, not the limit as delta u approaches zero. Now this right over here, just looking at it the way we can rewrite as du/dx, I think you see where this is going. this is u prime of x, or du/dx, so this right over here. But if u is differentiable at x, then this limit exists, and They're differentiable at x, that means they're continuous at x. So we assume, in orderįor this to be true, we're assuming. Order for this to even be true, we have to assume that u and y are differentiable at x. This is the definition, and if we're assuming, in So what does this simplify to? Well this right over here, So let me put some parentheses around it. the limit as delta x approaches zero, delta x approaches zero, of this business. Product of the limit, so this is going to be the same thing as the limit as delta x approaches zero of,Īnd I'll color-coat it, of this stuff, of delta y over delta u, times- maybe I'll put parentheses around it, times the limit. But what's this going to be equal to? What's this going to be equal to? Well the limit of the product is the same thing as the Would cancel with that, and you'd be left withĬhange in y over change x, which is exactly what we had here. Just going to be numbers here, so our change in u, this Change in y over change in u, times change in u over change in x. So I could rewrite this as delta y over delta u times delta u, whoops. I'm gonna essentially divide and multiply by a change in u. So this is going to be the same thing as the limit as delta x approaches zero, and I'm gonna rewrite Now we can do a little bit ofĪlgebraic manipulation here to introduce a change the derivative of y with respect to x, is equal to the limit asĭelta x approaches zero of change in y over change in x. Go about proving it? Well we just have to remind ourselves that the derivative of This with respect to x, we could write this as the derivative of y with respect to x, which is going to beĮqual to the derivative of y with respect to u, times the derivative This with respect to x, so we're gonna differentiate The derivative of this, so we want to differentiate So the chain rule tells us that if y is a function of u, which is a function of x, and we want to figure out Surprisingly straightforward, so let's just get to it, and this is just one of many proofs of the chain rule. Our independent variable, as that approaches zero, how the change in our function approaches zero, then this proof is actually And, if you've beenįollowing some of the videos on "differentiability implies continuity", and what happens to a continuous function as our change in x, if x is What I hope to do in this video is a proof of the famous and useful and somewhat elegant and
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