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Chain Rule

  1. May 1, 2007 #1
    All the proofs I have found for the Chain Rule involve limits and the fundamental theorem of Algebra...

    So I came up with a PROOF, not a derivation. But my teacher claims that my proof is invalid. Is it? If so, why???


    let:
    u=z(x)
    y=f(u)=f(z(x))

    therefore: dy/du = f ' (u)
    therefore: dy = (f ' (u)) * du -->

    therefore: du/dx = z ' (x)
    therefore: 1/dx = (z ' (x))/du -->


    therefore dy/dx = dy*(1/dx)
    substituting...
    therefore: dy/dx = ((f ' (u))*du)*(z ' (x))/du
    which simplifies to:
    dy/dx=(f ' (u))*(z ' (x))=(f ' (z(x)))*(z ' (x)) ==>
    or alternatively substituting...
    dy/dx=dy/du*du/dx ==>
     
  2. jcsd
  3. May 1, 2007 #2

    Office_Shredder

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    Uhh... is this rigorous? You're just manipulating notation. While extremely convenient, the use in a true proof is questionable.

    Of course a proof of the chain rule will involve limits.... because the definition of a derivative is based on a limit, and the chain rule is a proof about a derivative. You can often expect proofs to rely on definitions
     
  4. May 1, 2007 #3
    Your proof is valid. Though there is a simpler proof.

    du / dx = z'(x)

    df / du = f'(u)

    If we multiply both, we get df/dx, which is what we are looking for. Hence df / du = f'(u) * z'(x)
     
  5. May 1, 2007 #4
    Well, (to "Office Shredder"), the reason, I used arrows, was because my teacher could not follow my proof. Anyway, thanks. I just wanted to check whether my proof is valid. I insist that it is not a derivation. I see it like induction - this concept involves proving the statement/equation by proving that it will work for all the numbers defined within the set in which it claims that it will work... Although this is a simplistic thought, it underlies; the reason why I think my proof is valid. If I am "manipulating notation", then can you tell me why this is invalid in a 'Proof'???
     
  6. May 1, 2007 #5
    Manipulation of infinitesimals as such is valid because the meaning of the limit they represent is not lost.

    Edit: e.g. With the expression 1/dx = = (z ' (x))/du, the OP substituted such as dy/dx = ((f ' (u))*du)*(z ' (x))/du. This expression, interpreted as a limit, is still valid. Since du is a function of dx such as du = du(dx), if we let dx go to a very small value, so does du and the expression (z ' (x))/du becomes closer and closer to 1/dx. Since it gets closer and closer to 1/dx, ((f ' (u))*du)*(z ' (x))/du gets closer and closer to (f ' (u))*du / dx. Hence, if (f ' (u))*du / dx has a limit, the other expression will have the same limit because the two expressions become increasingly close as dx goes to 0.
     
    Last edited: May 2, 2007
  7. May 1, 2007 #6
    How exactly is this a proof? du/dx and df/du are not fractions.
     
  8. May 2, 2007 #7
    As said before, infinitesimals can be treated like numbers because the the ending result always represents the limit we are looking for once we "convert" this ending result to a limit. This said, the simpler proof can be understood in another way: as in limits, dy, df and dx are not 0. The expressions dy/dx and df/dx are thus fractions. Their factor, df/dx, is also a fraction. However, it is more convenient to look at df/dx as a product. The definition of the derivative is dy/dx = f'(x) + k where k is increasingly small for dx going to 0. Hence we would have

    df/dx = (z'(x) + k )*(f'(u) + l ) = z'(x)*f'(u) + lk + l(...) + k(...)

    You can see that as dx becomes closer to 0, the values l and k become very small, and so do the terms lk + l(...) + k(...). If we introduce a variable m, such as m = lk + l(...) + k(...), we obtain

    df/dx = z'(x)*f'(u) + m

    Since m can be made as small as we wish, this new expression fits the definition of the derivative and hence z'(x)*f'(u) is the derivative.
     
    Last edited: May 2, 2007
  9. May 2, 2007 #8
    Thanks. Your Proof to my proof is exactly what I asked for...
     
  10. May 2, 2007 #9

    Office_Shredder

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    werg, your definition of dy/dx doesn't mean anything as far as I can tell; can you clarify it?
     
  11. May 2, 2007 #10

    mathwonk

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    my stomach hurts. this is horse****
     
  12. May 2, 2007 #11
    I have no definition of dy/dx: I just assigned dy/dx to a fraction for the sake of being practical. dy/dx in my explanation really means f(x+h) - f(x)/h. Mathwonk, please elaborate...
     
  13. May 2, 2007 #12
    It doesn't seem like such a great idea, at least for rigorous proofs, to arbitrarily assign new definitions to things that are already well defined.

    But that is not what dy/dx is, it is the limit of this ratio.
     
  14. May 2, 2007 #13

    Office_Shredder

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    Then you hit the wall that

    [tex]lim_{h->0}\frac{f(x+h)-f(x)}{h}[/tex] =/= [tex]\frac{lim_{h->0}f(x+h)-f(x)}{lim_{h->0}h}[/tex]
     
  15. May 2, 2007 #14
    While I agree with the sentiment... not your most constructive post ever.

    (Nor mine :rolleyes: )
     
  16. May 2, 2007 #15

    Gib Z

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    Werg22, I am all for your treatment of the differentials in such a manner, but as a nice trick that use in calculations, not as a *proof*. If you wish to justify your treatment rigorously, please prove that in every case the differentials can be treated as such, retaining its original definition.
     
  17. May 2, 2007 #16
    Okay, the arguments seem to have swerved. But nobody has explained whether my PROOF (NOT derivation) is valid. If not, nobody has yet stated why...
     
  18. May 2, 2007 #17

    Gib Z

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    In The strictest sense, it is not valid as you have treated notation which just appears to look like a fraction, as a fraction. We have discussed the pros and cons and if you read our response you would have realised we have already stated why it is not valid.
     
  19. May 2, 2007 #18
    I don't think you got the whole principle. Forget dy/dx. Pretend we are talking about f(u+h) - f(u) / i

    where h is a small change change in u that is itself a function of the change in x, which we shall denote i.

    If we multiply the expression by the change h top and bottom, we get the fraction

    f(u+h) - f(u) / h * h/i

    Note that this a fraction and no limit has been evaluated. Now as i - > 0, so does h. But however small i and h, the expression simplifies to f(u+h) - f(u)/i.
    Hence, the limit as i goes to 0 of THAT expression, is the same as the limit of f(u+h) - f(u)/i as i goes to 0. We have (definition of a limit):

    f(u+h) - f(u) / h = f'(u) + k, h/i = z'(x) + l

    where k and l are increasingly small for smaller and smaller i and h. This gives us the following expression for f(u+h) - f(u) / i:

    f(u+h) - f(u) / i = f(u+h) - f(u) / h * h/i = f'(u)*z'(x) + kl + k(...) + l(...)

    Here again, I have only dealt with fractions. Now look at what happens: if we let i go to 0, so does h. Hence, for a very small i, k and l will be very small. Now you can see that the expression kl + k(...) + l(...) approaches 0 and is, from a certain point, constantly approaching 0 without being bounded to a value close to it. Hence we can write

    f(u+h) - f(u) / i = f'(u)*z'(x) + m(i)

    where m(i) is a function of i and is equal to kl + k(...) + l(...). Now since we have deduced that m(i) is increasingly small (absolute value) from a certain point and approaches 0, this new expression fits exactly what we mean by a limit. Hence f'(u)*z'(x) is defined as the limit. This is exactly what I did with my explanation that made dy/dx a real fraction: it was for the sake of being pragmatic. In the explanation, dy/dx were no longer infinitesimals, but just values such as f(u+h) - f(u), i and u. Now that I didn't use any of this, I hope it's clearer. Anyway, the real use of dy/dx spares us allot of time as it spares us to consider everything with limits. And for being rigorous, the explanation itself is plenty rigorous, just not the presentation - a ridiculous expectation on a forum.
     
    Last edited: May 2, 2007
  20. May 2, 2007 #19
    No wall at all, h is left to be the same on top and bottom so it's the same expression.
     
  21. May 2, 2007 #20

    Office_Shredder

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    Except the RHS of that inequality doesn't even exist.

    So you've proven no function has a derivative. Congratulations

    Notice how, when being rigorous, one can only prove lim(a/b) = lim(a)/lim(b) only if lim(a), lim(b) both exist and lim(b) =/= 0 (assuming lim(a/b) exists here)
     
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