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I have the following expression (everything is $\in \mathbb R$):

$f(a,b,c)=c\cdot\int_a^b g(t) \cdot h(t,c) \,dt,\quad0\leq a

I now want to differentiate this function with respect to c: $\frac{\delta f(\cdot)}{\delta c} $

I know how $h(\cdot)$ looks, but I have no definition of $g(t)$. Is there any way to get to the desired derivative without knowing $g(t)$?

If it is important, here is the definition of $h(\cdot)$:

$h(t,c)=e^{-t\cdot d\cdot(1-c)},\quad0


Edit: My original question has been answered super, now I wonder If there is also a solution if I whish to differentiate with respect to $a$ or $b$: $\frac{\delta f(\cdot)}{\delta a}$ As I understand it, the Leibniz rule can no longer be applied here, right?

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    @Steven Stadnicki: Ok, thanks.2011-04-28

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It looks like $h$ has continuous partials with respect to $t$ and $c$, so it's legal to use the Leibniz integral rule AKA "Differentiating under the integral sign". So here goes:

I'm guessing $c$ doesn't depend on $t$ and that all the functions involved are "nice".

You've got:

$ \begin{eqnarray*} \frac{\partial}{\partial c} c \int_a ^b g(t) h(t,c) dt &=& \int_a ^b g(t) h(t,c) dt + c \frac{\partial}{\partial c}\int_a ^b g(t) h(t,c) dt \\ &=& \int_a ^b g(t) h(t,c) dt + c \int_a ^b \frac{\partial}{\partial c}(g(t) h(t,c)) dt \\ &=& \int_a ^b g(t) h(t,c) dt + c \int_a ^b g(t) \frac{\partial}{\partial c}h(t,c) dt. \end{eqnarray*} $

Since $h$ is an exponential the derivative isn't so bad.

$ \begin{eqnarray*} \frac{\partial}{\partial c}h(t,c) &=& \frac{\partial}{\partial c}e^{-td + tdc} \\ &=& \frac{\partial}{\partial c}e^{-td}e^{tdc} \\ &=& e^{-td}te^{tc}= te^{-td + tdc} = tdh(t,c) . \end{eqnarray*}$

So at the end you get something like this:

$\begin{eqnarray*} &=& \int_a ^b g(t) h(t,c) dt + c \int_a ^b g(t) tdh(t,c) dt \\ &=& \int_a ^b (cdt+1)g(t) h(t,c) dt . \end{eqnarray*}$

There may be a sneaky way to evaluate this integral without knowing $g(t)$ but I'm in a rush now and not seeing it.

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    You seem to have copied the function $h$ slighty wrong (missing a $d$).2011-04-28
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The integrand satisfies the hypothesis of Leibniz's rule for differentiation under the integral sign. Don't worry about $g$, if it does not depend on $c$.