# Clarification Needed: Proof of the Product Rule

• Jan 1st 2013, 01:34 PM
Lambin
Clarification Needed: Proof of the Product Rule
Here's the proof, as provided by text:

$\frac{d}{dx}(uv)=\lim_{h\to\0}\frac{u(x+h)v(x+h)-u(x+h)v(x)+u(x+h)v(x)-u(x)v(x)}{h}$
$=\lim_{h\to\0}[u(x+h)\frac{v(x+h)-v(x)}{h}+v(x)\frac{u(x+h)-u(x)}{h}]$
$=\lim_{h\to\0}u(x+h)*\lim_{h\to\0}\frac{v(x+h)-v(x)}{h}+v(x)*\lim_{h\to\0}\frac{u(x+h)-u(x)}{h}.$

I understand that $u(v+h)v(x)-u(v+h)v(x)$ is introduced into the expression so that the equation can be expressed by definition as $\frac{dv}{dx}$ and $\frac{du}{dx}$. Although in step 2, wouldn't the denominator, h, retain its form into step 3 as:

$\lim_{h\to\0}\frac{u(x+h)}{h}*\lim_{h\to\0}\frac{v (x+h)-v(x)}{h}+\lim_{h\to\0}\frac{v(x)}{h}*\lim_{h\to\0} \frac{u(x+h)-u(x)}{h} ?$

It seems to me the h was omitted from a factor for each term.
• Jan 1st 2013, 01:42 PM
LordoftheFlies
Re: Clarification Needed: Proof of the Product Rule
Think about it: that is equivalent to saying:

$\dfrac{xy}{h}=\dfrac{x}{h}\cdot\dfrac{y}{h}=\dfrac {xy}{h^2}$

... which of course, is false!