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Why is this true?

  • Thread starter pivoxa15
  • Start date
  • #1
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Homework Statement


(d/dx)c=(d/da)c(da/dx)

where c=f(a(x))


The Attempt at a Solution


It seems correct because the da cancels but how do you get this from first principles?
 

Answers and Replies

  • #2
radou
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You may want to investigate the chain rule.
 
  • #3
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Could you possibly be a little bit more specific?
 
  • #5
HallsofIvy
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Homework Statement


(d/dx)c=(d/da)c(da/dx)

where c=f(a(x))


The Attempt at a Solution


It seems correct because the da cancels but how do you get this from first principles?
"da/dx" and "dc/da" are NOT fractions so it is not correct to say that the "da" cancels!

You may want to investigate the chain rule.
Could you possibly be a little bit more specific?
Are you saying you have never heard of the chain rule? You are being asked to prove the chain rule!
[tex]\frac{dc}{dx}= \lim_{h\rightarrow 0} \frac{c(x+h)- c(x)}{h}[/tex]
"Before" the limit, this is a fraction- you can cancel before the limit but be careful, this is not a trivial proof.
 
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  • #6
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I have heard of the chain rule.

In fact I realise that the chain rule is just what I stated (d/dx)c=(d/da)c(da/dx)
 
  • #7
2,255
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"da/dx" and "dc/da" are NOT fractions so it is not corret to say that the "da" cancels!
But physicists and applied mathematicians like to treat them as fractions in the limit. Is it okay to treat them as fractions and specify "in the limit"?
 
  • #8
HallsofIvy
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But physicists and applied mathematicians like to treat them as fractions in the limit. Is it okay to treat them as fractions and specify "in the limit"?
Yes, because the derivative is the limit of a fraction, you can always treat them "like" a fraction- that's one of the advantages of the dy/dt notation over f '. And, in fact, it is motivation for defining the "differentials", dx and dy= f '(x) dx.

My point here was that, if you are being asked to prove the chain rule, you can't just "treat dy/dx like a fraction" since the chain rule is part of what allows us to do that.
 
  • #9
Gib Z
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If i was a pedantic bastard, I would say perhaps pivoxa15 was speaking of using hyperreals? Thank god im not :D
 
  • #10
HallsofIvy
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If you do, I'll beat you around the head and shoulders with a 2 by 4!
 

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