What is the derivative of the real function $f:\mathbb{R}\rightarrow \mathbb{R}$ defined by $$f(x)=\int_{D(x)}g(\mathbf{t},x)\ d\mathbf{t},$$
where $D(x)=\{\mathbf{t}=(t_1,\ldots,t_n)\in \mathbb{R}^n: t_i \leq x,\ i=1,\ldots,n\}$ ?
I guess that under certain conditions the derivative is $$f'(x)=\int_{D(x)}\frac{d}{dx}g(\mathbf{t},x) \ d\mathbf{t}.$$
Am I correct ?
You have to consider also the "derivative" of $D(x)$ (shape calculus). If I'm correct, the result should be $$ f'(x) = \int_{D(x)} \frac{d}{dx} g(\boldsymbol t,x) \, d\boldsymbol t + \int_{\partial D(x)} g(\boldsymbol t, x) \, ds(\boldsymbol t). $$ Here, $s(\boldsymbol t)$ denotes the surface measure on $\partial D(x)$. Maybe you could prove that this is the correct derivative.
This is Reynolds transport theorem. I asked a question a few days ago about rigorous hypothesis, but no answer yet. Provided your $g$ and $D$ fullfill the required hypothesis, you have $$ \frac{df}{dx}(x) = \int_{D(x)}\frac{\partial g}{\partial x}({\bf t},x)d{\bf t} + \int_{\partial D(x)}g({\bf t},x)\frac{\partial \bf t}{\partial x}.{\bf n}_b d{\bf t}, $$ where ${\bf n}_b$ is the normal of the boudary $\partial D(x)$ of $D(x)$ at $\bf t$ and $\frac{\partial \bf t}{\partial x}$ depends on the parameterization you choose for this boundary.
Edit :
For instance, in your case, you can decompose the boundary $\partial D(x)$ into a set of hyperplanes $\mathcal{P}_k = \{{\bf t}=(t_1,\ldots,t_n)\in \mathbb{R}^n, t_k = x, t_i \leq x, i = 1,\ldots,n\}$, for $k = 1,\ldots,n$. Therefore, the boundary term in the formula becomes a sum over each of these hyperplanes. A parameterization of $\mathcal{P}_k$ is for instance the map $$ {\bf t} : \begin{array}{ccc} {\mathbb{R}^-}^{n-1} &\rightarrow& \mathcal{P}_k \\ (u_1,\ldots,u_{n-1}) &\mapsto& (x+u_1,\ldots,x+u_{k-1},x,x+u_{k},\ldots,x+u_{n-1}) \end{array} $$ The determinant of the Jacobian of this parameterization is $(-1)^k$. In addition, the normal of $\mathcal{P}_k$ consistent with the orientation of the parameterization we use is the vector $(-1)^ke_k$, where $e_k$ has coordinates $0$ except at the $k^{th}$ coordinate which is $1$. You also have $\frac{\mathrm{d} {\bf t}}{\mathrm{d} x} = (1,\ldots,1)$ and $\frac{\mathrm{d} {\bf t}}{\mathrm{d} x}.{\bf n}_b = (-1)^k$. Therefore you have $$ \int_{\mathcal{P}_k(x)} g({\bf t},x)\frac{\partial {\bf t}}{\partial x}.{\bf n}_b \mathrm{d}{\bf t} = \int_{-\infty}^0 \ldots \int_{-\infty}^0 g({\bf t}(u_1,\ldots,u_k),x) \mathrm{d}u_1\ldots\mathrm{d}u_{n-1} $$
