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Math Help - two Q in derivative Calculation by using prouduct rule

  1. #1
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    two Q in derivative Calculation by using prouduct rule

    Hi all

    Two Q in derivative Calculation by using prouduct rule

    see the attachment
    Attached Thumbnails Attached Thumbnails two Q in derivative Calculation by using prouduct rule-2121.jpg  
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  2. #2
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    EDIT: To clarify, I am answering part 2.

    When dealing with derivatives, the most important thing is to be careful. There are dropped and confused signs, arithmatic errors, and your substitution for v is wrong. (Also, not that it matters, but you kind of pulled your last step out of nowhere: continuing from the step above would give 2x). I'm sorry to say but I have to scrap everything except the mini-derivatives; those were correct.

    u'v + v'u =

    (1-\frac{1}{x^2})(x-\frac{1}{x}+1) \, \, + \, \, (1+\frac{1}{x^2})(x+\frac{1}{x})

     = (x-\frac{1}{x}+1) \, - \, (\frac{1}{x}-\frac{1}{x^3}+\frac{1}{x^2}) \, + \, (x+\frac{1}{x}) \, + \, (\frac{1}{x}+\frac{1}{x^3})

    x-\frac{2}{x}+1+\frac{1}{x^3}-\frac{1}{x^2}+x + \frac{2}{x} + \frac{1}{x^3}

    2x+1-\frac{1}{x^2}+\frac{2}{x^3}
    Last edited by Turiski; June 6th 2010 at 08:10 AM.
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  3. #3
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    Q1 should be +12x^2 instead of +6x^2

    Second last line of that answer you had -2x^2 and -x^2 instead of +
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  4. #4
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    Quote Originally Posted by r-soy View Post
    Hi all

    Two Q in derivative Calculation by using prouduct rule

    see the attachment
    1. You got the sign of x wrong. See attachment.

    2. y=\left(x+\frac1x  \right)\left(x-\frac1x +1  \right)

    Then

    y'=\left(1-\frac1{x^2}  \right)\left(x-\frac1x +1  \right) + \left(x+\frac1x  \right)\left(1+\frac1{x^2}  \right)

    Expand the brackets carefully, collect like terms. The final result contains 4 summands.
    Attached Thumbnails Attached Thumbnails two Q in derivative Calculation by using prouduct rule-zweifragen.png  
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  5. #5
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    Quote Originally Posted by Turiski View Post
    EDIT: To clarify, I am answering part 2.

    When dealing with derivatives, the most important thing is to be careful. There are dropped and confused signs, arithmatic errors, and your substitution for v is wrong. (Also, not that it matters, but you kind of pulled your last step out of nowhere: continuing from the step above would give 2x). I'm sorry to say but I have to scrap everything except the mini-derivatives; those were correct.

    u'v + v'u =

    (1-\frac{1}{x^2})(x-\frac{1}{x}+1) \, \, + \, \, (1+\frac{1}{x^2})(x+\frac{1}{x})

     = (x-\frac{1}{x}+1) \, - \, (\frac{1}{x}-\frac{1}{x^3}+\frac{1}{x^2}) \, + \, (x+\frac{1}{x}) \, + \, (\frac{1}{x}+\frac{1}{x^3})

    x-\frac{2}{x}+1+\frac{1}{x^3}-\frac{1}{x^2}+x + \frac{2}{x} + \frac{1}{x^3}

    2x+1-\frac{1}{x^2}+\frac{2}{x^3}

    in multimpling why you said (-1/x^2 ) (x) =
    -1/x not -1/x^-1


    ???
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  6. #6
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    Quote Originally Posted by r-soy View Post
    in multimpling why you said (-1/x^2 ) (x) =
    -1/x not -1/x^-1


    ???
    \frac{-1}{x^2}\cdot x = \frac{-x}{x^2} = \frac{-1}{x} \neq \frac{-1}{x^{-1}} = -x
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