- #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?

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- Thread starter pivoxa15
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- #1

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

where c=f(a(x))

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

- #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?

- #4

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Chain Rule: http://mathworld.wolfram.com/ChainRule.html

- #5

HallsofIvy

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"da/dx" and "dc/da" are NOT fractions so it is not correct to say that the "da" cancels!## 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?

You may want to investigate the chain rule.

Are you saying you have never heard of the chain rule? You are being asked toCould you possibly be a little bit more specific?

[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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In fact I realise that the chain rule is just what I stated (d/dx)c=(d/da)c(da/dx)

- #7

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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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Yes, because the derivative is theBut 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"?

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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- #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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