anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Owlfred
  • Owlfred
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anonymous
  • anonymous
i don't believe you. this is amazingly hard.
anonymous
  • anonymous
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

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anonymous
  • anonymous
what text are you using?
anonymous
  • anonymous
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
myininaya
  • myininaya
1 Attachment
anonymous
  • anonymous
very nice. not a proof though
anonymous
  • anonymous
what if h is a constant function? that is the problem with all these proofs
anonymous
  • anonymous
wait whats very nice but not a proof
anonymous
  • anonymous
brittT myinanaya has it, use that one
anonymous
  • anonymous
use what myinanaya wrote. it is good
anonymous
  • anonymous
that is very helpful!
myininaya
  • myininaya
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
myininaya
  • myininaya
oh by the way that is suppose to say deltax->0 not h->0 o nthat attachment
anonymous
  • anonymous
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
myininaya
  • myininaya
oh yeah f(g(0)) i don't think will work i understand
anonymous
  • anonymous
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
anonymous
  • anonymous
see serge lang calculus if you want a real proof. that is why i asked what text you were using. use myinanaya's proof.
myininaya
  • myininaya
or i mean it doesnt have to be zero but a constant yeah
anonymous
  • anonymous
http://www.mathnotes.org/index.php?pid=64#?pid=64
myininaya
  • myininaya
the first one is my proof yeah! lol
myininaya
  • myininaya
part of
anonymous
  • anonymous
i will be willing to bet cash that it is the proof in the text brittT is using
anonymous
  • anonymous
better explanation of what is wrong http://math.rice.edu/~cjd/chainrule.pdf
anonymous
  • anonymous
the flaw is that if \[g(x)=c\] a constant, then the denominator is identically 0
anonymous
  • anonymous
i am using what myininaya posted

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