![]() ![]() ![]() So we can rewrite this, as our change in u approaches zero, and when we rewrite it like that, well then this is just dy/du. As our change in x gets smallerĪnd smaller and smaller, our change in u is going to get smaller and smaller and smaller. we can rewrite this right over here, instead of saying delta x approaches zero, that's just going to have the effect, because u is differentiable at x, which means it's continuous at x, that means that delta u ![]() 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. Below this, we will use the chain rule formula method. In this example we will use the chain rule step-by-step. For example, differentiate (4 3) 5 using the chain rule. Multiply this by the derivative of the inner function. But if u is differentiable at x, then this limit exists, and To do the chain rule: Differentiate the outer function, keeping the inner function the same. 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 The rule remains the same, you just have to do it twice: differentiate the outermost function, keep the inside the same, then multiply by the derivative of the inside. This is a chain rule, within a chain rule problem. In practice, of course, you will need to use more than one of the rules we have developed to compute the derivative of a complicated function.- What I hope to do in this video is a proof of the famous and useful and somewhat elegant and Be careful - the only multiplication going on in that problem is the 'ax' part. This can be simplified of course, but we have done all the calculus, so that only algebra is left. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |