Is it allowed and if so, how to differentiate this integral? 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?
 A: 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.
A: 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 $c$.
