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anonymous

  • 5 years ago

i need to prove the chain rule for [f(g(h(x)))]=f'(g(h(x))g'(h(x))h'(x) using the limit definition of the derivative

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    i don't believe you. this is amazingly hard.

  3. anonymous
    • 5 years ago
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    in fact if you look in your text i am willing to bet they do not even prove the chain rule. at some point they will say "it is reasonable to believe that ..." and not actually prove it

  4. anonymous
    • 5 years ago
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    what text are you using?

  5. anonymous
    • 5 years ago
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    no its proven in my text but i do not understand how to add the third funtion into the given proof and im using calculus early transcendentals by jon rogawski

  6. myininaya
    • 5 years ago
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    1 Attachment
  7. anonymous
    • 5 years ago
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    very nice. not a proof though

  8. anonymous
    • 5 years ago
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    what if h is a constant function? that is the problem with all these proofs

  9. anonymous
    • 5 years ago
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    wait whats very nice but not a proof

  10. anonymous
    • 5 years ago
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    brittT myinanaya has it, use that one

  11. anonymous
    • 5 years ago
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    use what myinanaya wrote. it is good

  12. anonymous
    • 5 years ago
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    that is very helpful!

  13. myininaya
    • 5 years ago
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    satellitle like so you mean if h(x)=5 then no matter what change x happens h will always be the same so we have lim deltax->0 (5-5)/h=0/h=0

  14. myininaya
    • 5 years ago
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    oh by the way that is suppose to say deltax->0 not h->0 o nthat attachment

  15. anonymous
    • 5 years ago
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    well my problem also states that they are all differntiable and if it was a constant then it would not work and yeah i figured thats what you ment thank you

  16. myininaya
    • 5 years ago
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    oh yeah f(g(0)) i don't think will work i understand

  17. anonymous
    • 5 years ago
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    the problem with all these proof is they ignore what can happen if the "inside function" is a constant. but ignore me because you (brittT) clearly don't have to worry about it. forget i mentioned it. but a rigorous proof of the chain rule is a pain

  18. anonymous
    • 5 years ago
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    see serge lang calculus if you want a real proof. that is why i asked what text you were using. use myinanaya's proof.

  19. myininaya
    • 5 years ago
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    or i mean it doesnt have to be zero but a constant yeah

  20. anonymous
    • 5 years ago
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    http://www.mathnotes.org/index.php?pid=64#?pid=64

  21. myininaya
    • 5 years ago
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    the first one is my proof yeah! lol

  22. myininaya
    • 5 years ago
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    part of

  23. anonymous
    • 5 years ago
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    i will be willing to bet cash that it is the proof in the text brittT is using

  24. anonymous
    • 5 years ago
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    better explanation of what is wrong http://math.rice.edu/~cjd/chainrule.pdf

  25. anonymous
    • 5 years ago
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    the flaw is that if \[g(x)=c\] a constant, then the denominator is identically 0

  26. anonymous
    • 5 years ago
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    i am using what myininaya posted

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