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Math Help - one derivative and one integration

  1. #1
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    one derivative and one integration

    I am having trouble with this derivative:
    y = (2x+3/2x-3)^(1/4)
    The whole fraction is raised to the 1/4 power..I am not so sure where to start but I need to do logarithmic differentation to solve it.
    ln(y) = ln((2x+3/2x-3)^(1/4))

    The other problem I am having is this integration:
    (Integral sign) (xdx)/(x^4+3) The x is raised to just 4. the 3 is not an exponent.
    I substituted u to be equal to x^2 and then du = 2xdx.
    I now am stuck with 1/2(Integral sign) (du)/(u^2+3) The u is raised to 2..the 3 is not an exponent.

    Thanks for all the help!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by davecs77 View Post
    y = (2x+3/2x-3)^(1/4)
    The whole fraction is raised to the 1/4 power..I am not so sure where to start but I need to do logarithmic differentation to solve it.
    ln(y) = ln((2x+3/2x-3)^(1/4))
    You can do it by taking logarithms, but you don't need to. It's just a chain rule:
    y = f(g(x)) so y^{\prime} = f^{\prime}(g(x)) \cdot g^{\prime}(x)

    So in your case:
    y = \left ( \frac{2x + 3}{2x - 3} \right )^{1/4}

    y^{\prime} = \frac{1}{4} \left ( \frac{2x + 3}{2x - 3} \right )^{-3/4} \cdot \frac{2(2x - 3) - (2x + 3)(2)}{(2x - 3)^2}

    y^{\prime} = -3 \left ( \frac{(2x + 3)^{-3/4}}{(2x - 3)^{5/4}} \right )

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by davecs77 View Post
    y = (2x+3/2x-3)^(1/4)
    The whole fraction is raised to the 1/4 power..I am not so sure where to start but I need to do logarithmic differentation to solve it.
    ln(y) = ln((2x+3/2x-3)^(1/4))
    Doing it using logarithms,
    ln(y) = ln \left ( \left ( \frac{2x + 3}{2x - 3} \right )^{1/4} \right ) = \frac{1}{4} ln \left ( \frac{2x + 3}{2x - 3} \right ) = \frac{1}{4} ln(2x + 3) - \frac{1}{4} ln(2x - 3)

    Now take the derivative of both sides:
    \frac{1}{y} \frac{dy}{dx} = \frac{1}{4} \frac{1}{2x + 3} \cdot 2 - \frac{1}{4} \frac{1}{2x - 3} \cdot 2

    \frac{dy}{dx} = y \cdot \left ( \frac{1}{2(2x + 3)} - \frac{1}{2(2x - 3)} \right )

    Now sub in your given value of y and simplify. You should (if I did it right) get the same expression I got in the first post.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by davecs77 View Post
    The other problem I am having is this integration:
    (Integral sign) (xdx)/(x^4+3) The x is raised to just 4. the 3 is not an exponent.
    I substituted u to be equal to x^2 and then du = 2xdx.
    I now am stuck with 1/2(Integral sign) (du)/(u^2+3) The u is raised to 2..the 3 is not an exponent.
    Look at
    \int \frac{du}{u^2 + 3}

    Substitute u = \sqrt{3}z. Then
    \int \frac{du}{u^2 + 3} = \sqrt{3} \int \frac{dz}{3z^2 + 3} = \frac{\sqrt{3}}{3} \int \frac{dz}{z^2 + 1}

    Does this integral look more familiar? (Think trig substitution...)

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    Look at
    \int \frac{du}{u^2 + 3}
    \int\frac1{a^2+u^2}~du=\frac1{a}\arctan\frac{u}{a}  +k,~~a\ne0


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