# 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.

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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?

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Thanks you for your answer, it's helped clear a lot up, I have another question however, why will multiplying the derivatives give zero? –  seeker Apr 8 '13 at 16:02
@Assad When you multiply one really tiny quantity by another really tiny quantity, you get a really really tiny quantity. –  Baby Dragon Apr 8 '13 at 16:40
Another question what is the difference between ${{\delta y} \over {\delta x}}$ and ${{dy} \over {dx}}$, if I understood what you have said correctly, there is a difference between the two, could you expand on that? Thank you. –  seeker Apr 8 '13 at 18:14
@Assad: I was assuming some things about your notation that you didn't say explicitly. I was thinking of $\delta y$ as something like $(y + h) - y$ for some small $h$, or really as something like $( y(x+h) - y(x))$ for some small $h$. For $\frac{dy}{dx}$, I was thinking this referred to after one takes the limit as $h \to 0$. –  mixedmath Apr 8 '13 at 18:30

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:

$$d(U*V) = V*dU+U*dV$$ $$\int d(U*V) = \int V*dU+\int U*dV$$ $$UV = \int V*dU+\int U*dV$$ $$\int U*dV = UV - \int V*dU$$

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Please use the \begin{align}\end{align} environment together with the  delimiters to provide a more readable rendering of your answer. –  Lord_Farin Apr 8 '13 at 14:55