# Proof of the Product Rule

• April 23rd 2010, 01:52 PM
iPod
Proof of the Product Rule
I understand the product rule and how to use it, but I've been told to prove it, and I don't really know how. Could I be helped with this?
Thank you
• April 23rd 2010, 02:04 PM
pickslides
Shouldn't be too hard

Here's the start

$f(x) = v(x)\times u(x)$

Now by $f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

we have

$f'(x) = \lim_{h \to 0}\frac{v(x+h)u(x+h)-v(x)u(x)}{h}$

can you take it from here?
• April 23rd 2010, 02:08 PM
iPod
I have understood what you have done using the calculus theorem, however I don't know what to do from were you have left it, usually the ' $h$'s would cancel out,
Sorry, could you give more clues
• April 23rd 2010, 02:37 PM
pickslides
There's only a couple of steps left to the proof, I don't want to spoil it completely!

$f'(x) = \lim_{h \to 0}\frac{v(x+h)u(x+h)-v(x)u(x)}{h}$

here's a little bit more

$f'(x) = \lim_{h \to 0}\frac{v(x+h)u(x+h)-v(x+h)u(x)+v(x+h)u(x)-v(x)u(x)}{h}$

Now factor out $v(x+h)$ and $u(x)$ in groups.
• April 23rd 2010, 02:57 PM
iPod
I may have got it I think,

$\frac{v(x+h)u(x+h)-v(x+h)u(x)+v(x+h)u(x)-v(x)u(x)}{h}$

$= v(x+h)\frac{u(x+h)-u(x)}{h}+u(x)\frac{v(x+h)-v(x)}{h}$

but $\frac{u(x+h)-u(x)}{h}$ can be written as $u'(x)$ as h tends to zero

thus $f'(x)=v(x)u'(x)+u(x)v'(x)$ as h tends to zero

am I correct?
• April 23rd 2010, 03:02 PM
pickslides
(Nod)

Nice work!
• April 23rd 2010, 03:03 PM
iPod
wahay, kinda guessed that,
thank you very much :)