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Local maxima/minima and removable discontinuities
#12
OB3LISK Wrote:Dammit don't you nerds sleep. I came in about to dazzle everyone and all you pineappleing people are in here already. *MAD*

Gosh. Are you guys using the quotient rule when you're doing your derivative of (x^3)/x?? Or are you just doing the derivative of x^2...

If you do the quotient rule it turns out to be 3x - x...Which is what Kalovale got by bringing the x up to -1 power and using product rule.

Dammit 21whatshisface is right. The minimum is close to 0 but not 0.

That's quite interesting cause i don't spot that using the quotient rule. Though it's technically right.

dy/dx = [x*(3x^2)- (x^3)*1] / x^2 = [3x^3 - x^3] / x^2 = 2 x^3/x^2 = 2x. Or you can keep it as the fraction... either way.


Edit: Oh wait i see it... why would you split up the fraction though? xd
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Local maxima/minima and removable discontinuities - by shouri - 2011-03-24, 01:29 PM

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