Why is (du)(dv) negligible?

[Goldom]

"Here is Leibniz's argument: Let u(x) and v(x) be two differentiable functions of x. Then the differential of uv is

d(uv) = ((u+du)(v+dv)-uv) = (u(dv)+v(du)+(du)(dv))

Since the term (du)(dv) is "negligible", Leibniz concluded that

d(uv)=(du)v+u(dv)"

[LibertyorDeath]

1. Please lower your dosage

:)

[Droopy]

2. How about them Cardinals!

[4morewars]

3. Please UP your dosage !

:bounce:

[The Traveler]

4. Because

Something small multiplied by something small yields something very small.

No get your head out of the mathbooks and go watch some anime ie something.

:evilgrin:

[Goldom]

5. That's what I thought, but it isn't small, neccisarily.

ex. u(x)=5x^4, v(x)=10x^9, (du)=16x^3, (dv)=90x^8
(du)(dv)= 1440x^11.

That's not small at all.

[salib]

7. Yes it is!!!

Because it is in the limit as du and dv become small. Remember that anything less than 1 that is raised to a power greater than one is actually much smaller than 1. Thus, in your example, (du)(dv) is VERY small.

(Why am I doing this?)

[The Traveler]

8. I have had far

too many beers to contemplate this now and am as speechless as Bush when confronted by truth.

OK. I'LL watch anime! Back to the math books for you, my friend.

[MADem]

6. That Leibniz!!!!!!

He'll argue with anyone over anything!!!!

[salib]

9. Hey,

At least he was the first one who did not see fit to try to "justify" calculus with geometry examples.

[FDRrocks]

10. Why does 'i squared' = -1?

Man I suck at math, so I just .... follow the rules :)

Actually isn't it because i=the square root of 1. So then the question would be, why does i equal that.

[Bossy Monkey]

11. I can't imagine (Rimshot!) n/t