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

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- Apr 23rd 2010, 01:52 PMiPodProof 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 - Apr 23rd 2010, 02:04 PMpickslides
Shouldn't be too hard

Here's the start

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

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

we have

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

can you take it from here? - Apr 23rd 2010, 02:08 PMiPod
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 '$\displaystyle h$'s would cancel out,

Sorry, could you give more clues - Apr 23rd 2010, 02:37 PMpickslides
There's only a couple of steps left to the proof, I don't want to spoil it completely!

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

here's a little bit more

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

Now factor out $\displaystyle v(x+h)$ and $\displaystyle u(x)$ in groups. - Apr 23rd 2010, 02:57 PMiPod
I may have got it I think,

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

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

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

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

am I correct? - Apr 23rd 2010, 03:02 PMpickslides
(Nod)

Nice work! - Apr 23rd 2010, 03:03 PMiPod
wahay, kinda guessed that,

thank you very much :)