# Thread: proof of derivation product rule

1. ## proof of derivation product rule

I have question regarding to product rule.

We know that derivative of y with respect to x is:
$
\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:
$f'(x)\cdot g'(x) = f'(x)\cdot g(x) + f(x)\cdot g'(x)
$

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

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

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

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

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

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

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

2. Originally Posted by tabularasa
How i can get rid of $\lim_{h\rightarrow 0} \frac{\Delta u \cdot \Delta v}{h}$ since this is some error in my proof.
It is not an error. Write it as $\lim_{h\rightarrow 0} h\cdot\frac{\Delta u}h \cdot \frac{\Delta v}{h}$. Then $\lim_{h\to0}\frac{\Delta u}h = u'(x)$, $\lim_{h\to0}\frac{\Delta v}h = v'(x)$, and of course $\lim_{h\to0}h = 0$. Therefore, by the product rule for limits, $\lim_{h\rightarrow 0} h\cdot\frac{\Delta u}h \cdot \frac{\Delta v}{h} = 0$.

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

4. Originally Posted by tabularasa
Where's that h $\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.

5. 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!