 So today, I wanted to take a look at an intuitive proof for the chain rule that I found I think well, there is a more rigorous and this complex and more correct Proof, but It's unintuitive. It's a lot of algebraic manipulation and I I can follow the steps and see that it works, but I I Don't really I don't really feel the intuition there. So this this proof is more it's it's simpler and you just sort of get it You sort of get why why it's the case when you first start so We have to look a little deeper into what the derivative well first. Let me see what the channel is The channel basically states that if you want to find the derivative of a composite function f of g of x you have to that would be equal to the derivative of f with respect to g Multiplied by the derivative of g with respect to x Now We have to look we have to look at what the what what a derivative is so Basically when we're taking the derivative of function We're taking a tiny change in x in the input and a tiny change in the output and we're dividing the time change in the output by the tiny change in the input and Basically, we're getting a line Which is tangent Which is tangent to the to the function as a whole which is tangent to the function at that point x that you chose so When we're one so now that we know that really a derivative is just a tiny change in the output over a tiny change in the input we can we can say that The derivative of this is really df Over dx all right Now the hard part that comes with getting this is we have to account for g of x Which is just this middle step that sort of Messes with stuff, but that's a little easy so Again, we find the derivative or a tiny we take df over dg Which is this And then we multiply that By dg over dx Which is this? these two cancel out and Now we have df over dx and what always worked me when I initially saw the proof the chain rule not the proof is Why do we multiply by g prime of x? What does it even do and now after seeing this is I just see that it's a translation step by That makes it so that we're actually just taking df with respect to dx now Well, I'm just gonna take a look at two examples and they're not We're gonna take a look at two examples where we do sort of this This process so Let's take a factory that produces steel And takes in kilowatt hours as an input so we So for every kilowatt hour We take four k squared or And we make that number of tons of steel So we make four k squared tons of steel So every kilowatt hour mean we make that kilowatt hour squared times four tons of steel and Let's say we have a buyer beam and He buys steel in bulk. So the function that shows how much he pays for steel is 2000 times the square root of s which is tons of steel. So We want to know How much more money we make for every change in k for hum for how many more kilowatt hours we spend so or we use if We use them efficiently. So first we wanted to take that we want to find the change in B, of course So we take db that will equal 1000 Over the square root of s multiplied by ds Where was I'm so the change in B is 1000 over the square root of s times the change in s So we're not gonna we're not gonna turn this into db over ds For this reason so we want to find a change in s now And taking the we're just taking the derivative of that dk Now that we have this we can put this we can put this db all in terms of k so no need for s's so That would be 1000 over the square root of 4k squared because s is just 4k squared and we multiply that by ds which is 8k times dk If we divide db by dk That would equal let's see. That's 2k Over 1000 The case cancel that's 4,000 so db Over dk is 4,000 So for every change in k we make 4,000 extra dollars so That's our that's our derivative. I would do another example, but I think this sort of illustrates the point I'm not I'm not exactly sure how this measurement might be useful So I was planning on doing something like I don't know tell me in the comments if Another example of like Let's see if the same business if the same factory s Produces could produce copper and With a different resource And say w for worker so forever meter of copper wire That would take would be the number of workers squared plus So I'm just making this is just arbitrary and When we take a different a different buyer B, which buys copper wire and He buys it on a discounted rate on discounted, but he also buys it in bulk so he buys copper and For every meter of copper he buys 20 he's He does ten dollars times C Minus the square root of The copper Also times ten Plus ten as an overhead cost All right, so let's just simplify all of this to ten No, we should have we should have kept it as is Too late now. All right So of course first we got to find the the change in B. So DB For we drop out the constant and we get ten C and we do minus Five over the square root of C and we multiply all of this times the change in C And now we got to find DC, which is just to W Plus one so that's a number of workers. Oh Now we just replace all of this so Take ten W squared plus W minus five over the square root W squared plus W Multiply that by two W plus one and all that times DC and if we take DB over DC, we should cancel these out and this would be This whole expression here Would be a rate of change in our money per worker. So a profit Per doll the other thing the other thing was also the same Acceptance that of workers. There was a kilowatt hours. I think that's okay, man The comparison I was trying to make I'm not sure if this is accurate if this if this would be a good one But when does when does the advantage of adding more workers? Beat the advantage of adding more of having more kilowatt hours, so One point at what point does this? How many workers would it take for it to reach four thousand basically? I don't know if that's a I don't know if that's how it actually works if that's what it would actually represent But still I hope this video illustrated Why why it works and just sort of cleared up why? Multiplying by by the derivative of G is a translation. It's sort of like a translation step So, uh, yeah, thanks for watching You