# [SOLVED] Differentiation

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• January 25th 2009, 03:44 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
when I got c=1, I got it because I let -1=Left side $(y^2$+y). So I guess I should have done $(-1)^2$-(-1)= $\frac{(0)}{3}^3$+C. But this would give C to be 0

yes...and that's what C is

by the way, it is + (-1)

and if you learn to use LaTeX properly, you will realize you won't have to be chopping things up like that and putting only one term between each pair of [tex] tags
• January 25th 2009, 03:48 PM
ronaldo_07
and if you learn to use LaTeX properly, you will realize you won't have to be chopping things up like that and putting only one term between each pair of [tex] tags[/quote]

yea, sometimes error comes up when I try to put it all in 1 go
• January 25th 2009, 03:49 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
yea, sometimes error comes up when I try to put it all in 1 go

that's because you don't know how to use it properly :p, hence my last comment
• January 25th 2009, 04:02 PM
ronaldo_07
do I leave answer as y= $\frac{-1 \pm \sqrt{1+4x^3}}{2}$
• January 25th 2009, 04:17 PM
Jhevon
Quote:

Originally Posted by ronaldo_07
do I leave answer as y= $\frac{-1 \pm \sqrt{1+4x^3}}{2}$

well, it should be $\frac {4x^3}3$ instead of $4x^3$, but yes, that's the right form
• January 25th 2009, 04:24 PM
ronaldo_07
Quote:

Originally Posted by Jhevon
well, it should be $\frac {4x^3}3$ instead of $4x^3$, but yes, that's the right form

Thats what I got it was a typo, thanks for the help :)
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