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What's U? I see it in my textbook, but I don't understand.

@lgbasallote: this is not integration..

yeah...but i find it easier using u :C

Wait. Where'd that comment go. I was working on that...

derivative of x^-2 is -2x^-3

raised to -3 mimi :P have you forgotten your power rule /;) a^n = na^(n-1)

Answer in my textbook is y'=(-2(8x+3)/((4x^2+3x)^3)

\[y'=-2(8x+3)/(4x^2+3x)^3\]

yup seems about right

I don't know how to get that...

LOL you forgot the -2 this time mimi :PPPP

ahh that's about right :) do you get it now @IsTim ?

And sorry for the various typos..haven't touched derivivatives in a while..

lol, compare yoursteps to mine; which one is easier? :P

mine had 3 steps..yours had 3 reposts :p

just kdding dont kill me

lol, pity a poor kid who has not touched derivatives in a LONG time! :P

hahaha well considering as you have more medals..ill assume your method was easier :p

lol, only 1 medal difference dw. i was so nice that i gave you a free one. :p

that's the worst part mine was a pity medal :P

aww..i take sympathy to those who uses strange methods. :P

=))) it's innovation i tell you :p haha

Oh no. Mimi, there's a problem!

What's the problem?

The exponent of 3 on the denominator that surrounds 4x^2+3x. Where is it?

Woops, typo, lol..so much typos today sorry

i told you if you'd just use my method you wont miss that :P

haha missed it again :P

HAHAHAHHAA =))))

okay :C i just wanted to share my innovations

I don't understand. How?

Which step you do not understand?

remember how \(\large a^{-n} = \frac{1}{a^n}\)?

that is simple algebra..which i do not know how to explain..

No...

I don't remember.

Ok.

i.e. \(\LARGE a^{-n} = \frac{1}{a^n}\) and \(\Large \frac{1}{a^{-n}} = a^n\)