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Suppose that the first derivative of f(x) is g(x).
What the hell is the derivative of f(f(x))?
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On Mon, 16 Nov 2009 13:39:50 +0100, Invisible <voi### [at] dev null> wrote:
> Suppose that the first derivative of f(x) is g(x).
>
> What the hell is the derivative of f(f(x))?
http://www.google.com/search?q=What+is+the+derivative+of+a+composite+function%3F
http://en.wikipedia.org/wiki/Chain_rule
Basic high-school level math. I am slightly surprised that you did not
know it, though I am not in the least surprised that you did not bother to
google for it.
--
FE
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Fredrik Eriksson wrote:
> Basic high-school level math. I am slightly surprised that you did not
> know it,
What, you mean given that I've never been tought mathematics?
> though I am not in the least surprised that you did not bother
> to google for it.
Oh, I did Google it. I just didn't comprehend the answer. (In
particular, I've already read the Wikipedia entry.)
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On Mon, 16 Nov 2009 13:51:11 +0100, Invisible <voi### [at] devnull> wrote:
>
> Oh, I did Google it. I just didn't comprehend the answer. (In
> particular, I've already read the Wikipedia entry.)
What is so difficult to understand about it? I thought function
composition would be one of the areas of math that you were actually good
at, given your fanatical obsession with a certain functional programming
language.
Anyway:
The derivative of f( g(x) ) is f'(g(x))*g'(x).
Substitute the inner function into the derivative of the outer function,
and then multiply by the derivative of the inner function.
In your example, we want the derivative of f(f(x)) (i.e. the outer and
inner functions are the same), and we know that f'(x) = g(x) (try not to
mix up all the f's and g's).
df(f(x))/dx = f'(f(x))*f'(x) = g(f(x))*g(x)
--
FE
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>> Oh, I did Google it. I just didn't comprehend the answer. (In
>> particular, I've already read the Wikipedia entry.)
>
> What is so difficult to understand about it?
One formula appears to say the derivative is f(x) * g(x), and another
appears to say it's f'(g(x)) * g'(x), which is different, but these are
supposed to be different notations for the same formula...?
> I thought function
> composition would be one of the areas of math that you were actually
> good at, given your fanatical obsession with a certain functional
> programming language.
Function composition I get. It's what that does to the derivating I
can't quite wrap my brain around.
> Anyway:
>
> The derivative of f( g(x) ) is f'(g(x))*g'(x).
>
> Substitute the inner function into the derivative of the outer function,
> and then multiply by the derivative of the inner function.
> df(f(x))/dx = f'(f(x))*f'(x) = g(f(x))*g(x)
Right. So basically, the function I'm looking for is f'(f(x))*f'(x)?
So presumably the derivative of f(f(f(x))) is going to be something like
f'(f(f(x))) * f'(f(x)) * f'(x)? And in general, the Nth composition will
be ever more complex as N increases?
Oh dear.
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Fredrik Eriksson wrote:
> Basic high-school level math. I am slightly surprised that you did not
> know it, though I am not in the least surprised that you did not bother
> to google for it.
...what I *should* have done of course is
http://www.wolframalpha.com/input/?i=derivative+f(f(x))
*facepalm*
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Invisible wrote:
> So presumably the derivative of f(f(f(x))) is going to be something like
> f'(f(f(x))) * f'(f(x)) * f'(x)? And in general, the Nth composition will
> be ever more complex as N increases?
>
> Oh dear.
Ooo, but wait... The Nth derivative of a polynomial will be zero if N
exceeds the order of the polynomial. Hmm.
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>> So presumably the derivative of f(f(f(x))) is going to be something
>> like f'(f(f(x))) * f'(f(x)) * f'(x)? And in general, the Nth
>> composition will be ever more complex as N increases?
>>
>> Oh dear.
>
> Ooo, but wait... The Nth derivative of a polynomial will be zero if N
> exceeds the order of the polynomial. Hmm.
...which doesn't help, because the derivative of the Nth iterate will be
f'(x) * f'(f(x)) * f'(f(f(x)) * f'(f(f(f(x))) * ...
which involves only the first derivating of f and the iterates of f.
Damnit, I really suck at math. :'{
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Invisible wrote:
> Damnit, I really suck at math. :'{
D f[n](x) = D f[n-1](x) * f'(f[n-1](x))
Maybe this isn't so bad...
Or maybe I just need to GROW A FRIGGIN' BRAIN! >_<
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On Mon, 16 Nov 2009 14:14:27 +0100, Invisible <voi### [at] devnull> wrote:
>
> One formula appears to say the derivative is f(x) * g(x), and another
> appears to say it's f'(g(x)) * g'(x), which is different, but these are
> supposed to be different notations for the same formula...?
The first one is obviously wrong, unless you mixed up all the f's, g's and
little circles. It is difficult to tell without context.
>> df(f(x))/dx = f'(f(x))*f'(x) = g(f(x))*g(x)
>
> Right. So basically, the function I'm looking for is f'(f(x))*f'(x)?
Yes.
> And in general, the Nth composition will be ever more complex as N
> increases?
Yes.
> ...what I *should* have done of course is
> http://www.wolframalpha.com/input/?i=derivative+f(f(x))
Except that it is the wrong answer, or rather the wrong question. Wolfram
thought you meant (f <dot> f)(x), when what you actually meant was (f
<circle> f)(x).
Just one more reason why it is better to learn the formulas yourself
instead of relying on a computer to help you all the time.
--
FE
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