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Math Help - ln :o

  1. #1
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    ln :o

    <br />
y = cos (ln x)<br />

    im thinking:

    <br />
y' = (-sin) \cdot \frac{1}{x}<br />

    help thanks
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  2. #2
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    skeeter's Avatar
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    try again ...

    y = \cos(u)

    note that u is a function of x

    y' = -sin(u) \cdot \frac{du}{dx}
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  3. #3
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    so the answer is:

    <br /> <br />
y' = -sin(ln x) \cdot \frac {1}{x}<br />
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