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Math Help - Calculus help

  1. #1
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    Calculus help

    Hey guys,iv just started doing calculus and im struggling to simplify these 2 equations any further,can any1 help me?

    1. [ 2e^-w multiplied by(1+e^-3w) -2e^-w multiplied by (3e^-3) ] divided by [(1+e^-3)^2]


    2. [ 2cos(2Q)(inv3Q) + (3sin2q /3q) ]


    thanks guys
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  2. #2
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    Quote Originally Posted by Ritchie187 View Post
    2. [ 2cos(2Q)(inv3Q) + (3sin2q /3q) ]
    First off the variables Q and q are technically different. Are they supposed to be the same?

    And what function is "inv?"

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ritchie187 View Post
    1. [ 2e^-w multiplied by(1+e^-3w) -2e^-w multiplied by (3e^-3) ] divided by [(1+e^-3)^2]
    Unless the e^{-3} terms are supposed to be e^{-3w} all I can think to do is factor a e^{-w} out of the numerator., then add the constant terms in the numerator.

    -Dan
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    sory, the q's are suposed to be the same and Inv was suposed to be In for inverse. thanks
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ritchie187 View Post
    sory, the q's are suposed to be the same and Inv was suposed to be In for inverse. thanks
    To be clear: Inv{3Q} = 1/(3Q)?

    -Dan
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  6. #6
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    the only bit i realy need to simplify in question 2 is the last bit i think, simplify

    (3sin2q) divided by 3q. i think the answer should either be, (sin2q)/q or just sin2? what do you think?
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    Quote Originally Posted by Ritchie187 View Post
    the only bit i realy need to simplify in question 2 is the last bit i think, simplify

    (3sin2q) divided by 3q. i think the answer should either be, (sin2q)/q or just sin2? what do you think?
    First you should be writing it as
    [3sin(2q)]/(3q)

    Now, it is NOT true that sin(2q) = 2sin(q) or any other such manipulation: the sine function is NOT linear!!

    The best you can do is:
    [3sin(2q)]/(3q) = [sin(2q)]/q

    -Dan
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