$\frac{d}{d x_1} \int_{-c}^{F(x_1,x_2)}v_1(x_1,x_2,x_3)dx_3\overset{!}{=}-v_1(x_1,x_2,F(x_1,x_2))\frac{\partial}{\partial x_1}F(x_1,x_2)$

Question

This equation is part of a proof of the divergence theorem in $\mathbb{R}^3$ for sets $$M=\bigg\{x \in \mathbb{R}^3: x_3 \leq F(x_1,x_2) \bigg\} \cap[0,1]^2\times[-c,c]$$ in which $F: [0,1]^2 \rightarrow [-c,c]$ is differentiable and $v: \mathbb{R}^3 \rightarrow \mathbb{R}^3, v_{|\partial ([0,1]^2\times[-c,c])}\equiv0 $ is the vector-field: $$\frac{d}{d x_1} \int_{-c}^{F(x_1,x_2)}v_1(x_1,x_2,x_3)dx_3=-v_1(x_1,x_2,F(x_1,x_2))\frac{\partial}{\partial x_1}F(x_1,x_2)$$ Using chain rule i get: $$\frac{d}{d x_1} \int_{-c}^{F(x_1,x_2)}v_1(x_1,x_2,x_3)dx_3$$$$=\left ( \frac{d}{d x_1} V_1 \right )(x_1,x_2,F(x_1,x_2))-\left ( \frac{d}{d x_1} V_1 \right )(x_1,x_2,-c)+v_1(x_1,x_2,F(x_1,x_2))\frac{\partial}{\partial x_1}F(x_1,x_2)$$ Here $\frac{d}{dx_1}V_1=v_1$. There brackets are there to indicate, that only $V_1$ is differentiated and then concatinated with $f(x_1,x_2)=(x_1,x_2,F(x_1,x_2))$ not the other way around. There has to be something added and subtracted in between, but I have no idea, what exactly that could be.