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Hi,
I have a question regarding differentiation:
I am under strict instruction that the chain rule may not be used - at least not without proof from first principles.
Many thanks
Last edited by Au101 (2011-11-01 10:08:06)
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Hi;
"From first principles" to me means the delta method. Is that your impression too?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi bobbym - I wasn't sure what it meant - my friend told me that they had been set this problem and thought I might be interested. I'm curious, but couldn't get anywhere with it myself, so I thought I'd ask here. I tried to do it with limits, using the traditional definition of a derivative as:
I suppose I was essentially trying to get around not being able to simply use the chain rule, by getting to the result of its application, but this may well not be the best way to do it.
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Hi;
Last time I did this was nearly a century ago so it might not be rigorous enough for todays treatment of an infinitesimal.
Now subtract line 2 from line 4.
Do you follow what happened from line 5 to line 6?
Essentially it is an algebraic method of finding the derivative.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi bobbym, thanks a lot, but I think I'm missing a bit of theory - perhaps I'm just a bit rusty, lol. I've missed working on my mathematics, but I've recently started a university course in Sanskrit and Linguistics and haven't had the free time .
Anyway, I'm not really sure why:
And also, I don't know why, in the proceeding lines - having expanded the brackets, we are left with the extra y and x.
Thanks for all your help
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Hi;
In any function when we peturb or add a little increment called Δx to x then you know that the dependent variable y must also change by something. We call that change Δy. We do not know what Δy or Δx is, just that they are very, very small.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Ah yes, of course, you're quite right, sorry, I was just thinking about it the wrong way, I read it as adding something to the y and then the same to the x, but i see what's going on now, but what about when we expand the brackets?
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What part?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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And then of course, this is repeated further on. I'm sure I will see it the moment you tell me, I'm fairly sure I'm just not approaching this from the right perspective, but i don't quite see why we have the y and the x after the expansion, and then lose the a-squared.
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Hi;
Just algebra, like multiplying two binomials.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I'm sorry bobbym, I think tonight is a bad time for me, but since I've put you to the effort of helping me, let's see if I can get my head around this. Delta denotes change in y, that's fine, that's old hat, so a small change in y multiplied by a small change in y is a small (albeit larger, but that's not really relevant) change in y-squared. So let's think about the expansion:
Okay, sorry, I think I'm with you? I just completely misinterpreted the notation. Yes, the rest of it now makes sense, I think my sleep-deprived state was holding me back - thank you very much for your answer and your patience
Last edited by Au101 (2011-11-01 12:27:30)
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Hi;
and then lose the a-squared.
The a^2 is gone due to the subtraction of the second equation from the fourth equation.
You can think of it in a better way. Not delta x which means nothing to anyone but as a little bit of something.
This little concept which was first expressed by Silvanus Thomson in his book "Calculus Made Easy," ( this book taught zillions of us dummies how to do calculus ) makes the next concept that
much easier to understand and apply.
so a small change in y multiplied by a small change in y is a small (albeit larger, but that's not really relevant)
This statement is not correct. If it were the move made in line 5 to line 6 would not be possible. This is important.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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