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.

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?
