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Thread: 3 derivative questions!

  1. #1
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    3 derivative questions!

    i have no clue how to do this one:
    $\displaystyle

    y = (ln \cdot e^\frac{1}{x} \cdot cos^2 \cdot \sqrt {x})
    $

    this one i tried:
    $\displaystyle

    y = ln (e^3 \cdot \sqrt {x^2 + 2x} \cdot \sqrt {sin2x})
    $
    and the answer i got was:
    $\displaystyle

    y' = \frac {3(x^2+2x) + e(x+1) + e(x^2+2x)(tan2x)}{e(x^2+2x)}
    $

    and this one i have no idea either
    $\displaystyle

    y = log_3 (x^2 + 3x + 5)
    $

    THANKS!!!
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  2. #2
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    Quote Originally Posted by qzno View Post
    i have no clue how to do this one:
    $\displaystyle

    y = (ln \cdot e^\frac{1}{x} \cdot cos^2 \cdot \sqrt {x})
    $

    this one i tried:
    $\displaystyle

    y = ln (e^3 \cdot \sqrt {x^2 + 2x} \cdot \sqrt {sin2x})
    $
    and the answer i got was:
    $\displaystyle

    y' = \frac {3(x^2+2x) + e(x+1) + e(x^2+2x)(tan2x)}{e(x^2+2x)}
    $

    and this one i have no idea either
    $\displaystyle

    y = log_3 (x^2 + 3x + 5)
    $

    THANKS!!!
    $\displaystyle y = \log_3 (x^2 + 3x + 5)$

    $\displaystyle 3^y = x^2 + 3x + 5$

    $\displaystyle \ln{3^y} = \ln{x^2 + 3x + 5}$

    $\displaystyle y \ln{3} = \ln{x^2 + 3x + 5}$

    $\displaystyle y = \frac{1}{\ln{3}} \ln{x^2 + 3x + 5}$

    Let $\displaystyle u = x^2 + 3x + 5$ so $\displaystyle y = \frac{1}{\ln{3}} \ln{u}$

    $\displaystyle \frac{du}{dx} = 2x + 3, \frac{dy}{du} = \frac{1}{\ln{3}} \frac{1}{u} = \frac{1}{(x^2 + 3x + 5)\ln{3}}$

    $\displaystyle \frac{dy}{dx} = \frac{2x + 3}{(x^2 + 3x + 5)\ln{3}}$
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  3. #3
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    Joined
    Oct 2008
    Posts
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    thanks hahah but i figured that one out after

    $\displaystyle

    y = (ln \cdot e^\frac{1}{x} \cdot cos^2 \cdot \sqrt {x})
    $

    what about this one?
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