I want to prove that $dV(x,y,z)=-\vec{E(x,y,z)} \cdot\vec{ds} \iff V(x,y,z)=-\int_{(x,y,z)}^{\infty} \vec{E(x,y,z)} \cdot \vec{ds}$. I keep losing minus sign, I can't find error myself. Every step seems right, although answer is wrong.
$dV=-\vec{E} \cdot\vec{ds} \implies V=-\int_{(x,y,z)}^{\infty} \vec{E} \cdot \vec{ds}$ is self evident.
$dV(x,y,z)=\frac{\partial V(x,y,z)}{\partial x} dx+\frac{\partial V(x,y,z)}{\partial y} dy+\frac{\partial V(x,y,z)}{\partial z} dz$
$V(x,y,z)=-\int_{(x,y,z)}^{\infty} \vec{E} \cdot \vec{ds} \iff V(x,y,z)=\int_{\infty}^{(x,y,z)} \vec{E} \cdot \vec{ds}$
$V(x,y,z)=\int_{\infty}^{x} {E_x(x,y,z)} dx+\int_{\infty}^{y} {E_y(x,y,z)} dy+ \int_{\infty}^{z} {E_z(x,y,z)} dz$ $dV(x,y,z)=\frac{\partial\int_{\infty}^{x} {E_x(x,y,z)} dx}{\partial x}dx+\frac{\partial\int_{\infty}^{y} {E_y(x,y,z)} dy}{\partial y}dy+ \frac{\partial\int_{\infty}^{z} {E_z(x,y,z)} dz}{\partial x}dz+$ $dV(x,y,z)={E_x(x,y,z)} dx+{E_y(x,y,z)} dy+ {E_z(x,y,z)} dz$
$dV(x,y,z)=\vec{E(x,y,z)\cdot \vec{ds}}$
Where is the error?