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composition of functions


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Assuming all the function are well-behaved (i.e. all the derivatives exist) you just take the derivative with respect to x straight forwardly.


I'm going to drop the (t) notation -- it is still in there, x is still a function of t, I'm just not going to write it all out.


[math]\frac{d}{dx} f = \frac{d}{dx} \left(\frac{dx}{dt} +x\right)[/math]

[math]\frac{d}{dx} f = \frac{d}{dx}\frac{dx}{dt}+\frac{d}{dx}x[/math]


now, this is where the "well-behaved" comes in, the order of the differentiation with respect to x & t can be reversed:


[math]\frac{d}{dx} f = \frac{d}{dt}\frac{dx}{dx}+\frac{d}{dx}x[/math]

[math]\frac{d}{dx} f = \frac{d}{dt}1+1[/math]

[math]\frac{d}{dx} f = 1[/math]


I have no idea what Snail is talking about with matrices and the like. What matrix? where? Furthermore, what is there to "solve"? I really don't get any of Snail's reply.

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I was replying to a completely different forum and question, not sure how that happened....sorry about that. :embarass:


That makes a lot more sense -- I can move posts around the math section here, but I don't think they've given me enough power to move posts across forums (...yet).

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Bignose, under what conditions is




I tried to do some examples, say:


[math] x = t^2 [/math]


[math] \frac{dx}{dt} = 2t [/math] [math]\Rightarrow[/math] [math] \frac{d}{dx}\frac{dx}{dt} = \frac{1}{\sqrt{x}}[/math] since [math] \sqrt{x}=t [/math]




[math] \frac{d}{dt}\frac{dx}{dx} = 0 [/math]


So [math] \frac{d}{dt}\frac{dx}{dx}\neq \frac{d}{dx}\frac{dx}{dt} [/math]

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I'm not ignoring your question, I just haven't had a lot of time to get back to it. My first instinct is that because your f(x) actually isn't a function (which maps a single number to a single number) but is actually a functional (which maps a function to a function), that the definition of differentiation of the function isn't the quite right.


see http://mathworld.wolfram.com/FunctionalDerivative.html


I need some more time to look at this, so I'll get back when I've thought about it more...

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