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

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

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 $h(\cdot)$:

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


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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@Stijn: $g(t)$ is a yet to define growth term, e.g. linear or exponential. So I can easily make the assumption or claim that it is both. –  meep.meep Apr 28 '11 at 14:54
3  
For the question in your edit, you don't need anything as high-powered as Leibnitz - all you need there is the fundamental theorem of calculus... –  Steven Stadnicki Apr 28 '11 at 15:41
    
@Steven Stadnicki: Ok, thanks. –  meep.meep Apr 28 '11 at 16:21

2 Answers 2

up vote 3 down vote accepted

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 $d$). –  Raeder Apr 28 '11 at 15:02
    
Thanks! Fixed it. –  Schmitty Apr 28 '11 at 16:11

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 $c$.

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