A general proof of the Chain Rule Let f and g be differentiabl...

Question

A general proof of the Chain Rule Let f and g be differentiable functions with h(x)=f(g(x)). For a given constant a, let u=g(a) and v=g(x), and define H (v) = <1x1 matrix>

c. Show that h′(a) = lim x→a ((H(g(x))+f′(g(a)))⋅g(x)−g(a)/x−a).

Accepted Answer

Welcome back, everyone. In this problem, giving the differentiable functions R and S with N X equals R S X, and for a specific constant C, let W equals of C and Z equals S X with K Z equal to R Z minus RW divided by Z minus W if minus R W if Z is not equal to W and 0 if Z is equal to W. Determine the expression for N of C. For our answer choices, A says it's the limit as X approaches C of K S X plus R S C multiplied by S X plus SC all divided by X + C. B says it's the limit as X approaches C of K S X minus R S C multiplied by S X minus S C divided by X minus C. C says it's the limit as X approaches C of K S X plus R S C multiplied by S X minus S C divided by X minus C and the D says it's the limit as X approaches C of K S X plus R S C multiplied by S C minus S X divided by X minus C. Now how are we going to determine our expression here for the derivative of N of C, that is N of C? Well, we can use the different quotient formula for the derivative. Recall that it basically tells us that the derivative of NFC. OK. Is going to be equal to the limit. As X approaches C of N X minus enough C all divided by X minus C. So if we're going to apply the difference quotient here to get an expression for the derivative of NFC, that means we'll need to find enough X and enough C and express them in known terms based on our definition for K of Z. So, let's try to do that. What can we do? Well, first of all, remember in our problem, we were told that NFX equals RFS X, OK. So, given that we have a definition for NF X, OK. Which is R S X. And that would also mean then that NFC. Would be equal to R of SFC, OK. Then by the difference quotient formula. This tells us that the derivative of NFC is going to be equal to the limit as X approaches C of RFS X. Minus R of SFC. And there's enough X and enough C. Divided by X minus C. No, we still have some more work to do because we'll need to figure out what R of SF X is and RF SFC is, OK? So how can we do that? Well, again, remember we were given a definition that incorporates our function R for our K of Z, OK? So now, we know that based on the definition for K of Z, that means K of X, OK? K S X, using the definition for K of Z where Z is not equal to W, then K of X S X is going to be equal to R of S of X, OK, minus R of S of C. Let me close this here. OK. All divided by SF X. Minus SFC. So basically, we're just putting SFX and SFC where Z and W are, OK, minus the derivative of R of SFC, replacing the derivative of R of W. OK, R W. No, in this case, remember, we're trying to get an expression for R S X minus R of SFC. So we can do that by adding R of SC to both sides of our equation and then multiplying both sides by S X minus S of C, OK? In other words, We're going to be left with K of S of X, OK. Plus our of SFC, OK, we've added R of SFC to both sides, multiplied, OK. Bye. The denominators of X minus S of C. And all of this is gonna be equal to R of S of X. Minus R of as of C. Know that we have an expression for RF SFX minus R of SFC, we can substitute it into our equation for the derivative of NFC, OK? So, no, that means then. That the derivative of NRC is going to be the limit, as X approaches C of what we had for R S X minus RF SFC, this entire expression. OK, so that's K S X. Plus our prime of SRC OK. Multiplied By SFX. Minus SFC. OK. And all of this is gonna be divided where we can just write it over here for our form. So this is gonna be divided by X minus C. Therefore, this is the derivative of NFC. If we look back at our answer choices, then that tells us that C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

A general proof of the Chain Rule Let f and g be differentiable functions with h(x)=f(g(x)). For a given constant a, let u=g(a) and v=g(x), and define H (v) = <1x1 matrix>c. Show that h′(a) = lim x→a ((H(g(x))+f′(g(a)))⋅g(x)−g(a)/x−a).

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A general proof of the Chain Rule Let f and g be differentiable functions with h(x)=f(g(x)). For a given constant a, let u=g(a) and v=g(x), and define H (v) = <1x1 matrix>
c. Show that h′(a) = lim x→a ((H(g(x))+f′(g(a)))⋅g(x)−g(a)/x−a).