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Math Help - [SOLVED] Differentiation

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ronaldo_07 View Post
    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
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  2. #17
    Member ronaldo_07's Avatar
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    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
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  3. #18
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ronaldo_07 View Post
    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 , hence my last comment
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  4. #19
    Member ronaldo_07's Avatar
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    do I leave answer as y= \frac{-1 \pm \sqrt{1+4x^3}}{2}
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  5. #20
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ronaldo_07 View Post
    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
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  6. #21
    Member ronaldo_07's Avatar
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    Quote Originally Posted by Jhevon View Post
    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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