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Math Help - First and Second Derivative with ln, Help please

  1. #1
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    First and Second Derivative with ln, Help please

    Hey, I am having trouble finding either the first or second derivative of a problem. Could someone show me how to solve both and simplify? It will be really helpful with my other homework problems.


    f(x)= x(ln x)^2

    I need to find f '(x) and f ''(x). Thank you
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  2. #2
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    Quote Originally Posted by lostsheep View Post
    Hey, I am having trouble finding either the first or second derivative of a problem. Could someone show me how to solve both and simplify? It will be really helpful with my other homework problems.


    f(x)= x(ln x)^2

    I need to find f '(x) and f ''(x). Thank you
    First the product rule, then the chain rule.

    f(x) = x \cdot ln^2(x)

    f^{\prime}(x) = (x)^{\prime} \cdot (ln^2(x)) + x \cdot (ln^2(x))^{\prime}

    Now,
    (ln^2(x))^{\prime} = 2 \cdot ln(x) \cdot \frac{1}{x}, so
    f^{\prime}(x) = 1 \cdot (ln^2(x)) + x \cdot 2 \cdot \frac{ln(x)}{x}

    f^{\prime}(x) = ln^2(x) + 2ln(x)

    Then
    f^{\prime \prime}(x) = 2 \cdot ln(x) \cdot \frac{1}{x} + 2 \cdot \frac{1}{x}

    f^{\prime \prime}(x) = \frac{2ln(x)}{x} + \frac{2}{x}

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    First the product rule, then the chain rule.

    f(x) = x \cdot ln^2(x)

    f^{\prime}(x) = (x)^{\prime} \cdot (ln^2(x)) + x \cdot (ln^2(x))^{\prime}

    Now,
    (ln^2(x))^{\prime} = 2 \cdot ln(x) \cdot \frac{1}{x}, so
    f^{\prime}(x) = 1 \cdot (ln^2(x)) + x \cdot 2 \cdot \frac{ln(x)}{x}

    f^{\prime}(x) = ln^2(x) + 2ln(x)

    Then
    f^{\prime \prime}(x) = 2 \cdot ln(x) \cdot \frac{1}{x} + 2 \cdot \frac{1}{x}

    f^{\prime \prime}(x) = \frac{2ln(x)}{x} + \frac{2}{x}

    -Dan
    in the original problem it is stated f(x) = x(ln x)^2

    so (ln x) part is squared, is this the same as ln^2 (x)?

    I am not sure, but thanks for the help
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  4. #4
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    Quote Originally Posted by lostsheep View Post
    in the original problem it is stated f(x) = x(ln x)^2

    so (ln x) part is squared, is this the same as ln^2 (x)?

    I am not sure, but thanks for the help
    Yes they are the same thing with different notations,
    just as sin^2(x)=(sinx)^2.
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