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Given a function $c(r)$, where $r \in [0,1]$ and another function

$$ x = \alpha \times c(0) $$

where $\alpha$ is a constant and $c(0)$ is the function $c$ evaluated at $r = 0$, my question is - what is $\frac{dx}{dc}?$

As $c(r)$ varies, $c(0)$ will vary as well and so I assume the answer is $\frac{dx}{dc}$= $\alpha \frac{d c(0)}{d c}$. But if so, how do I evaluate $\frac{d c(0)}{d c}$?

Thanks!

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    $x$ only depends on the value of $c$ at the origin, so you need to carefully define what it is that you mean by $\frac{\mathrm{d} x}{\mathrm{d} c}$.2012-08-27
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    By $\frac{dx}{dc}$ I mean the derivative of x to the function $c(r)$ - defined everywhere not just at the origin. I agree that x only depends on the value of c at the origin - does that mean that $\frac{dx}{dc} = \frac{dx}{d c(0)}$?2012-08-27
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    The question needs really needs more context to give a satisfactory answer, but the you need to view the term $\frac{d c(0)}{d x}$ as a linear map from whatever space $c$ lies in to the codomain of $c$. So, in this case, you must have $\frac{d c(0)}{d x}(h) = h(0)$, if this is a continuous map (and makes sense).2012-08-27
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    @ilp The derivative is defined as a limit of differences. Could you write such differences for the derivative you have in mind?2012-08-27
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    $\frac{dx}{dc} = \lim_{\Delta c(r) \rightarrow 0} \frac{ x(c(r) + \Delta c(r) ) - x(c(r)) }{ \Delta c(r) } $2012-08-27
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    You need to have some norm (or similar) here for the term $\Delta c(r) \to 0$ to make sense.2012-08-27
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    True, $\frac{dx}{dc} = \lim_{|\Delta c(r)|_{2} \rightarrow 0} \frac{ x(c(r) + \Delta c(r) ) - x(c(r)) }{ \Delta c(r) } $2012-08-27
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    I don't think the derivative is defined unless $c$ is restricted to a certain family of functions.2012-08-27
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    @copper.hat I'm not sure I follow about $\frac{dc(0)}{dx}$ being a linear map from whatever space c lies in to the codomain of c. Can you explain?2012-08-27
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    There are various sorts of derivatives when dealing with points in 'function space'. The Frechet derivative is, in a prescribed sense, a best linear approximation to the function at a point. For example, with $f(x) = x^2$, we have eg, $f'(1) = 2$, hence the derivative at $1$ is the function $g : \mathbb{R} \to \mathbb{R}$ given by $g(h) = f'(1) h$, but we identify such derivatives with the number $f'(1)$ rather than the equivalent linear operator $g$. The difference is almost irrelevant for functions $\mathbb{R} \to \mathbb{R}$, but becomes significant when we go beyond finite dimensions.2012-08-27
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    (I ran out of space.) So here we have the function $f(c) = c(0)$, hence the derivative **if it exists** must be given by (since $f$ is linear) $g(h) = h(0)$, that is, the function that takes a function $h$ and returns the value at $0$. However, it is a function now, not just multiplication by a number.2012-08-27
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    @ copper.hat Thanks - that's beginning to make more sense now. I am still unclear about how you extended $f(x) = x^2, g(h) = f'(x) h$ to the case where $f(c) = c(0), g(h) = f'(c) h = h(0)$ ? In the case where $g(h) = h(0)$ what is the relationship between h and c?2012-08-27
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    For $f(x) = x^2$, the derivative is the function $g(h) = f'(1)h$, for the function $f(x) = x(0)$, the derivative is $g(h) = h(0)$. I wasn't extending as such, I was just pointing out that in the real case, the derivative is a linear operator (except we just think of it as a number).2012-08-27
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    Oops, I missed a point: In the case where $f(c) = c(0)$, since $f$ is linear, the derivative is the same as the function itself, so $c$ doesn't show up anywhere. If the function was not linear, for example $f(c) = c(0)^2$, then the derivative at $c$ would be $g(h) = 2c(0)h(0)$.2012-08-27

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