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    May I please have the proof for the chin rule as it dosnt seem to be in the book, may i please have proof for these different variants of the chain rule

    " \bullet if  y= [f(x)]^n then  \frac{dy}{dx} =n[f(x)]^n^-^1 f'(x) .

     \bullet if y= f[g(x)] then \frac{dy}{dx}=f'[g(x)]g'(x).

     \bullet [tex] \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} ."

    I may be able to understand the last two if i get a proof for the first.

    Thanks.
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    Its all the same thing gogle chain rule should help explain it as long as you know first principles
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    (Original post by thecakeisalie1)
    Its all the same thing gogle chain rule should help explain it as long as you know first principles
    The first principles? Of the chain rule or differentiation, i wonder what i was thinking when i spelt it as defrentiation in the thread title.
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    The proof of the chain rule isn't really A-level standard, it's quite a bit harder to understand, so for A-level you really just have to accept that the result is true. The second two results are just the chain rule (they're the same thing), and the first result is just an application of it (with f(u) = u^n and g(x) relabelled as f(x)).

    To show that the second two are the same thing, suppose that u=g(x) and y=f(u) = f(g(x)). Then \dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} = f'(u) g'(x) = f'(g(x))g'(x).
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    (Original post by nuodai)
    The proof of the chain rule isn't really A-level standard, it's quite a bit harder to understand, so for A-level you really just have to accept that the result is true. The second two results are just the chain rule (they're the same thing), and the first result is just an application of it (with f(u) = u^n and g(x) relabelled as f(x)).

    To show that the second two are the same thing, suppose that u=g(x) and y=f(u) = f(g(x)). Then \dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} = f'(u) g'(x) = f'(g(x))g'(x).
    Ah i see now, ok im going to keep reading this and think it over in my head until it fully sinks in but im begining to understand, although im not glad that they want me to use this without getting the proof for it, il just have to accept and understand for the sake of my grades. thanks, BTW i havent been allowed to rate for the past week is it something to do with your account?
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    (Original post by Core)
    Ah i see now, ok im going to keep reading this and think it over in my head until it fully sinks in but im begining to understand, although im not glad that they want me to use this without getting the proof for it, il just have to accept and understand for the sake of my grades. thanks, BTW i havent been allowed to rate for the past week is it something to do with your account?
    If you've +repped me recently then you won't be able to do it again for a while.

    To be honest, there's a lot of stuff in A-level which you just have to accept without it being proved, it's just that the chain rule more obviously falls into this category than other things. For example, it's never proved that you can differentiate almost any of the functions you come across (except perhaps constant functions, x and x^2 if you're lucky), and yet I'm sure you differentiate functions without too much hesitation -- you're just told that you can do it and you get on with it. There are many many more things too but I won't list them because you'll just be disappointed :p:
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    (Original post by nuodai)
    If you've +repped me recently then you won't be able to do it again for a while.

    To be honest, there's a lot of stuff in A-level which you just have to accept without it being proved, it's just that the chain rule more obviously falls into this category than other things. For example, it's never proved that you can differentiate almost any of the functions you come across (except perhaps constant functions, x and x^2 if you're lucky), and yet I'm sure you differentiate functions without too much hesitation -- you're just told that you can do it and you get on with it. There are many many more things too but I won't list them because you'll just be disappointed :p:
    I think the differention of polynomials is proved in c2, but i have come across polynomials with terms that i have not seen before so yes i guess your right there, oh and i also just wanted to say that after looking over your example a few more times i understood the chain rule(maybe not the proof but the logic behind its form) so thanks alot, and i haven't been able repped you in awhile, its strange.

    I tried to make a proof of it myself using binomial expansion but it's late and would have taken awhile (without any gurantee that i wouldn't be left more confused), I was also meant to have finished c3 by the 25th of march but I found the trig to be tough. Thanks again for your help on that also.

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