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.
 A: The basic idea here is this: Suppose $y(x) = u(x)v(x)$, so that a little change in $y$ comes from a little change in $u$ and a little change in $v$. We're going to look at a ratio of a little change in $y$ over a little change in $x$. As this $x$ change gets really, really small, this ratio (which we'd been denoting by $\frac{\delta y}{\delta x}$ approaches the derivative $\frac{dy}{dx}$, and similarly for the derivatives of $u$ and $v$.
But what about the $\frac{\delta u \delta v}{\delta x}$ term? The idea is that as the change in $x$ gets really small, both $\delta v$ and $\delta x$ get really small. While one ratio, say $\frac{\delta u}{\delta x}$ approaches a finite derivative, multiplying by the other will go to zero.
So that's what the 'proof' is about. Does that make sense?
A: 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:
$$d(U*V) = V*dU+U*dV$$
$$\int d(U*V) = \int V*dU+\int U*dV$$
$$UV = \int V*dU+\int U*dV$$
$$\int U*dV = UV - \int V*dU$$
