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# Chain rule proof

## Video transcript

what I hope to do in this video is a proof of the famous and useful and somewhat elegant and sometimes infamous chain rule and if you've been following some of the videos on differentiability implies continuity and what happens to a continuous function is our change in X if X is our independent variable as that gets down as that approaches zero how the change in our function approaches zero then this proof is actually surprisingly straightforward so let's just get to it this is just one of many proofs of the chain rule so the chain rule tells us that if Y Y is a function of U which is a function of X and we want to figure out the derivative of this so we want to differentiate this with respect to X so we're going to differentiate this with respect to X we could write this as the derivative of Y with respect to X which is going to be equal to the derivative of Y with respect to U times the derivative of U with respect to X this is what the chain rule tells us but how do we actually go about proving it well we just have to remind ourselves that the derivative of Y with respect to X the derivative of Y with respect to X is equal to the limit as Delta X approaches zero of change in Y over change in X now we can do a little bit of algebraic manipulation here to introduce a change in U so let's do that so this is going to be the same thing as the limit as Delta X approaches zero and I'm going to rewrite this part right over here I'm going to essentially divide and multiply by a change in U so I can rewrite this as Delta Y over Delta u times Delta u ty UPS times Delta u over Delta X change in Y over change in u times changing u over change in X and you can see this is just going to be numbers here so our change in you this would cancel with that and you'd be left with change in Y over change in X which is exactly what we had here so nothing nothing earth-shattering just yet 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 product of the limits so this is going to be the same thing as the limit as Delta X approaches 0 of colour-coded of this stuff of Delta Y over Delta u times maybe I'll put parentheses around it x times the limit the limit as Delta X approaches 0 Delta X approaches 0 of this business so I put some parentheses around it Delta u over Delta X so what does this simplify to well this right over here this is the definition and we're assuming in order for this to even be true we have to assume that U and Y are differentiable at X so we assume in order for this to be true we're assuming we're assuming y comma u are differentiable are differentiable are differentiable at X and remember also if they're differentiable at X that means they're continuous in X but if u is differentiable at X then this limit exists and this is the derivative of this is u prime of X or d u DX so this right over here we can rewrite as d u DX I think you see where this is going now this right over here just looking at it the way this written it-- right here we can't quite yet call this dy D u because this is the limit as Delta X approaches 0 not the limit as Delta u approaches 0 but we just have to remind ourselves the results from probably the previous video depending on how your watching it which is if we have a function you that is continuous at a point that as Delta X approaches 0 Delta u approaches 0 so we can actually rewrite this we can wreak an rewrite this right over here instead of saying Delta X approaches 0 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 is going to approach zero as our change in X gets smaller and smaller and smaller our change in U is going to get smaller and smaller and smaller so we can rewrite this as our change in U approaches 0 and when we rewrite it like that well then this is just dy D U this is just dy the derivative of Y with respect to U so just like that if we assume Y is and u are differentiable at X or you could say that Y is a function of U which is a function of X we've just shown in fairly simple algebra here and using some assumptions about differentiability and continuity that it is indeed the case that the derivative of Y with respect to X is equal to the derivative of the Y with respect to U times the derivative of U with respect to X hopefully you find that convincing
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