Partial derivative of integral function Let $h=h(x,t): \mathbb{R}\times [0,+\infty)\longmapsto \mathbb{R}$ and suppose $h,h_x \in C(\mathbb{R}\times [0,+\infty))$. 
Consider the function $$F(x,t)=\int_{0}^{t}h(x-c(t-s),s)ds,$$ where $c$ is a constant.
Is it true that $F(x,t)\in C^1(\mathbb{R}\times [0,+\infty))$?
In particular, how can I show that $F$ is continuous on $\mathbb{R}\times [0,+\infty)$? Does there exist a generalization of Leibniz integral rule (in the present case, the integrand depends on two parameters x,t) to calculate $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial t}$?
Thanks in advance!
 A: PERSONAL ATTEMPT
I think a generalization of Leibniz integral rule can be something like this
"Let $f=f(x,t,s)$ be a continuous function in some region $[x_0,x_1]\times[t_0,t_1]\times[s_0,s_1]$ of the $(x,t,s)$ space. Let $\alpha=\alpha(x,t)$ and $\beta=\beta(x,t)$ be continuously differentiable functions on $[x_0,x_1]\times[t_0,t_1]$ such that $s_0\leq\alpha(x,t)\leq\beta(x,t)\leq s_1\,\,\forall (x,t) \in [x_0,x_1]\times[t_0,t_1]$. 
Then $$F=F(x,t)=\int_{\alpha(x,t)}^{\beta(x,t)}f(x,t,s)ds$$ is continuous on $[x_0,x_1]\times[t_0,t_1]$.
Moreover, if partial derivatives of $f$ with respect to $x$ and $t$, $\frac{\partial f(x,t,s)}{\partial x} \,\text{and}\, \frac{\partial f(x,t,s)}{\partial t}$,    exist and are continuous on $[x_0,x_1]\times[t_0,t_1]\times[s_0,s_1]$ then $F=F(x,t)\in C^1([x_0,x_1]\times[t_0,t_1])$ and $$\frac{\partial F(x,t)}{\partial x}=\int_{\alpha(x,t)}^{\beta(x,t)}\frac{\partial f(x,t,s)}{\partial x}ds+ f(x,t,\beta(x,t))\cdot \frac{\partial \beta(x,t)}{\partial x}- f(x,t,\alpha(x,t))\cdot \frac{\partial \alpha(x,t)}{\partial x},$$ $$\frac{\partial F(x,t)}{\partial t}=\int_{\alpha(x,t)}^{\beta(x,t)}\frac{\partial f(x,t,s)}{\partial t}ds+ f(x,t,\beta(x,t))\cdot \frac{\partial \beta(x,t)}{\partial t}- f(x,t,\alpha(x,t))\cdot \frac{\partial \alpha(x,t)}{\partial t}$$ for all $(x,t)\in [x_0,x_1]\times[t_0,t_1]$."
