Differentiating a multivariable integral I am trying to analyze a function $F$ of $k$ variables, $x_1,x_2,\cdots ,x_k$. Specifically, if we let 
$$G(x_1,x_2,\cdots,x_k) = \int_{0}^{b(x_1,x_2,\cdots,x_k)}F(y,x_2,\cdots,x_k)dy,$$
is there any way I can express either $\frac{dG}{dx_i}$ or $\frac{\partial G}{\partial x_i}$ in terms of the derivatives of $F$ and $b$? I am aware of Leibniz's rule for differentiating under the integral sign, but that would still leave expressions with integrals of partial derivatives, and if possible I would love to learn about some technique that would allow me to express these derivatives just in terms of $F$, $b$ and their derivatives. 
I'm hopeful that there is a way to do this because the integration is only with respect to one of the variables of $F$. We can assume that all the applicable integrals and derivatives exist. 
 A: Denoting $H(t, x_2, \dots, x_n)= \int_0^t F(y, x_2, \dots, x_n) \ dy$, we have
$$ G(x_1, \dots, x_n) = H(b(x_1, \dots, x_n), x_2, \dots, x_n).$$
Then for $i \ge 2$, applying chain rule, we get:
$$\frac{\partial G}{\partial x_i}=\frac{\partial b}{\partial x_i}\frac{\partial H}{\partial t} + \frac{\partial H}{\partial x_i}$$
The fundamental theorem of calculus leads to
$$\frac{\partial H}{\partial t}(t, x_2, \dots, x_n) = F(t, x_2, \dots, x_n)$$
and derivative under integral sign to 
$$\frac{\partial H}{\partial x_i}(t, x_2, \dots, x_n) = \int_0^t\frac{\partial F}{\partial x_i}(y, \dots, x_n) \ dy$$
Summarizing all we finally get:
$$\begin{aligned}
\frac{\partial G}{\partial x_i}(x_1, x_2, \dots, x_n)&= \frac{\partial b}{\partial x_i}(x_1, x_2, \dots, x_n)F(b(x_1,x_2, \dots, x_n), x_2, \dots, x_n)\\
&+ \int_0^{b(x_1, x_2, \dots, x_n)}\frac{\partial F}{\partial x_i}(y, \dots, x_n) \ dy
\end{aligned}$$
The $x_1$-partial derivative is simpler and equal to
$$\frac{\partial G}{\partial x_1}(x_1, x_2, \dots, x_n)= \frac{\partial b}{\partial x_1}(x_1, x_2, \dots, x_n)F(b(x_1,x_2, \dots, x_n), x_2, \dots, x_n) dy$$
