# proof of derivation product rule

• Nov 17th 2008, 03:54 AM
tabularasa
proof of derivation product rule
I have question regarding to product rule.

We know that derivative of y with respect to x is:
$\displaystyle \frac{dy}{dx} = f'(x) = \lim_{h\rightarrow 0} \: \frac{f(x+h) - f(x)}{x+h-x} = \lim_{h\rightarrow 0} \: \frac{f(x+h) - f(x)}{h} = \frac{\Delta y}{\Delta x}$

Now I'm trying to understand the proof of product rule:
$\displaystyle f'(x)\cdot g'(x) = f'(x)\cdot g(x) + f(x)\cdot g'(x)$

to simplify notation:
$\displaystyle let\;u=f(x)\;and\;v=g(x).\;\;Then \;D(u\cdot v) = u' \cdot v + u \cdot v'$

Ok, now to proof:
$\displaystyle (u \cdot v)' = \lim_{h\rightarrow 0} \frac{\left[u(x+h) \cdot v(x+h) \right] - \left[u(x) \cdot v(x) \right]}{h}$

we know that $\displaystyle \Delta u = u(x+h) - u(x) \;,\; thus\; u(x+h) = \Delta u + u(x)$ so we can substitute this with respected to u and v in equation and get:
$\displaystyle (u \cdot v)' = \lim_{h\rightarrow 0} \frac{\left[(\Delta u+u(x)) \cdot (\Delta v + v(x)) \right] - \left[u(x) \cdot v(x) \right]}{h}$

Now to do some algebra we get:
$\displaystyle = \lim_{h\rightarrow 0} \left\{ \frac{\Delta u \cdot \Delta v}{h} + \frac{\Delta u}{h} \cdot v(x) + \frac{\Delta v}{h} \cdot u(x) + \frac{u(x) \cdot v(x)}{h} - \frac{u(x) \cdot v(x)}{h} \right\}$

$\displaystyle = \lim_{h\rightarrow 0} \left\{ \frac{\Delta u \cdot \Delta v}{h} + \frac{\Delta u}{h} \cdot v(x) + \frac{\Delta v}{h} \cdot u(x) \right\}$

And we know that $\displaystyle u' = \lim_{h\rightarrow 0} \frac{\Delta u}{h} \; , \; v' = \lim_{h\rightarrow 0} \frac{\Delta v}{h}$

so finally we substitute u' and v' to the equation we get:
$\displaystyle = \lim_{h\rightarrow 0} \frac{\Delta u \cdot \Delta v}{h} + u' \cdot v + v' \cdot u$

How i can get rid of $\displaystyle \lim_{h\rightarrow 0} \frac{\Delta u \cdot \Delta v}{h}$ since this is some error in my proof.
• Nov 17th 2008, 05:07 AM
Opalg
Quote:

Originally Posted by tabularasa
How i can get rid of $\displaystyle \lim_{h\rightarrow 0} \frac{\Delta u \cdot \Delta v}{h}$ since this is some error in my proof.

It is not an error. Write it as $\displaystyle \lim_{h\rightarrow 0} h\cdot\frac{\Delta u}h \cdot \frac{\Delta v}{h}$. Then $\displaystyle \lim_{h\to0}\frac{\Delta u}h = u'(x)$, $\displaystyle \lim_{h\to0}\frac{\Delta v}h = v'(x)$, and of course $\displaystyle \lim_{h\to0}h = 0$. Therefore, by the product rule for limits, $\displaystyle \lim_{h\rightarrow 0} h\cdot\frac{\Delta u}h \cdot \frac{\Delta v}{h} = 0$.
• Nov 17th 2008, 06:08 AM
tabularasa
Quote:

Originally Posted by Opalg
It is not an error. Write it as $\displaystyle \lim_{h\rightarrow 0} h\cdot\frac{\Delta u}h \cdot \frac{\Delta v}{h}$. Then $\displaystyle \lim_{h\to0}\frac{\Delta u}h = u'(x)$, $\displaystyle \lim_{h\to0}\frac{\Delta v}h = v'(x)$, and of course $\displaystyle \lim_{h\to0}h = 0$. Therefore, by the product rule for limits, $\displaystyle \lim_{h\rightarrow 0} h\cdot\frac{\Delta u}h \cdot \frac{\Delta v}{h} = 0$.

Where's that h $\displaystyle \lim_{h\to0} h\cdot\$ coming from?
• Nov 17th 2008, 06:23 AM
FusionHK
Quote:

Originally Posted by tabularasa
Where's that h $\displaystyle \lim_{h\to0} h\cdot\$ coming from?

Notice that he split the denominator into h^2, so there has to be another h in the numerator to counter that.
• Nov 17th 2008, 06:29 AM
tabularasa
Quote:

Originally Posted by FusionHK
Notice that he split the denominator into h^2, so there has to be another h in the numerator to counter that.

Aah, too clever. i just didn't saw that one coming. Thank you!