Cannot understand this part of the product rule for differentiation proof This is a proof of the product formulae for differentiation in my text book:
$\eqalign{
  & y + \delta y = (u + du)(v + \delta v) \to (1)  \cr 
  & y = uv \to (2) \cr} $
Subtracting equation 1 from 2:
$\eqalign{& \delta y = (u + \delta u)(v + \delta v) - uv  \cr 
  &  = uv + u\delta v + v\delta u + \delta u\delta v - uv  \cr 
  &  = \upsilon \delta v + v\delta u + \delta u\delta v  \cr 
  & {{\delta y} \over {\delta x}} = u{{\delta v} \over {\delta x}} + v{{\delta u} \over {\delta x}} + {{\delta u} \over {\delta x}}\delta v \cr} $
This is the part I do not understand, it says:
As $\delta x \to 0$ then ${{\delta y} \over {\delta x}} \to {{dy} \over {dx}}$, ${{\delta u} \over {\delta x}} \to {{du} \over {dx}}$ and ${{\delta v} \over {\delta x}} \to {{dv} \over {dx}}$
Also $\delta v \to 0$ and thus ${{\delta u} \over {\delta x}}\delta v \to 0$
Therefore:
${{dy} \over {dx}} = u{{dv} \over {dx}} + v{{du} \over {dx}}$

I'm wondering what is essentially being said in this portion of the proof, could you please explain in layman's terms, this is extremely confusing as it is.

Try this one. Just applying definition of derivative to $U(x)*V(x)$, no words at all:
$$y(x)=U(x)*V(x)$$ 
$$ y'(x) =\lim_{\delta x \to 0}{\space}{[y(x+\delta x)-y(x)]/\delta x}=$$
$$\begin{align}
\lim_{\delta x \to 0}{\space}{U(x+\delta x)*V(x+\delta x)-U(x)*V(x)/\delta x} 
=\lim_{\delta x \to 0}{\space}{U(x+\delta x)*V(x+\delta x)-U(x)*V(x)+U(x)*V(x+\delta x)-U(x)*V(x+\delta x)/\delta x} 
=\lim_{\delta x \to 0}{\space}{U(x+\delta x)*V(x+\delta x)-U(x)*V(x+\delta 
x)+U(x)*V(x+\delta x)-U(x)*V(x)/\delta x}  \end{align}$$
$$= U(x)'V(x)+V(x)'U(x)$$
Now we can easy get integration by parts formula just by integrating last equation:
$$d(U*V) = V*dU+U*dV$$
$$\int d(U*V) = \int V*dU+\int U*dV$$
$$UV = \int V*dU+\int U*dV$$
$$\int U*dV = UV - \int V*dU$$
