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Math Help - Derivatives and logarithm

  1. #1
    Newbie
    Joined
    Jun 2010
    Posts
    9

    Derivatives and logarithm

    I have this question :

    Find out derivative : y = √(x √(x + sineⁿ))

    { I use : n = x }

    I named : u = x + sine

    so,

    y' = (x + √y) (1 + u) u'

    { I use : a = - }

    And : u' = 1 + (sine)'

    But I have many problems with the last number. I Think it would be something like this :

    Ln y = x Ln e
    1/y y' = x/sine sine' + Ln sine
    y'/y = (x cose + sine Ln sine) / sine
    y' = sineⁿ‾ (x cose + sine Ln sine)

    Finally I think there is something wrong, cause at last, I find out a very complex expression...

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  2. #2
    Banned
    Joined
    Jan 2010
    From
    Tehran
    Posts
    33
    using the bernoulli's method is much faster .

    1 . Take the Ln from two sides.
    2 . Now : Derivinting from two sides :

    get  e ^ n = \theta

    Ln(y) = \frac{1}{2} Ln( x\sqrt{x+sin\theta}) =>

    Ln(y) = \frac{1}{2}[Ln(x) +Ln( \sqrt{x+sin\theta})] = >

    \frac{y'}{y} = \frac{1}{2}( \frac{1}{x} + \frac{1+cos\theta}{2\sqrt{x+sin\theta}}) =>

    y' = y \frac{1}{2}( \frac{1}{x} + \frac{1+cos\theta}{2\sqrt{x+sin\theta}}) =>

    y' = \sqrt{x\sqrt{x+sin\theta}}( \frac{1}{x} + \frac{1+cos\theta}{2\sqrt{x+sin\theta}}) \frac{1}{2} =>

    set  \theta = e ^ n  = >

    y' = \sqrt{x\sqrt{x+sin(e ^ n)}}( \frac{1}{x} + \frac{1+cos(e ^ n)}{2\sqrt{x+sin(e ^ n)}}) \frac{1}{2}
    Last edited by parkhid; June 4th 2010 at 12:37 AM.
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