Results 1 to 3 of 3

Thread: ln :o

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    109

    ln :o

    $\displaystyle
    y = cos (ln x)
    $

    im thinking:

    $\displaystyle
    y' = (-sin) \cdot \frac{1}{x}
    $

    help thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3701
    try again ...

    $\displaystyle y = \cos(u)$

    note that u is a function of x

    $\displaystyle y' = -sin(u) \cdot \frac{du}{dx}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2008
    Posts
    109
    so the answer is:

    $\displaystyle

    y' = -sin(ln x) \cdot \frac {1}{x}
    $
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum