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Math Help - Determining dy/du, du/dx, dy/dx

  1. #1
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    Determining dy/du, du/dx, dy/dx

    y={\sqrt u} + \frac {1}{\sqrt u}

    I found out what \frac{dy}{du} is and what \frac{du}{dx} and when I do \frac{dy}{dx} I get \frac {(3x^2)(x^3-1)}{2(x^3)^\frac {3}{2}} is this correct? or is it just \frac {(3x^2)(x^3)}{2(x^3)^\frac {3}{2}}
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  2. #2
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    Quote Originally Posted by oregon88 View Post
    y={\sqrt u} + \frac {1}{\sqrt u}

    I found out what \frac{dy}{du} is and what \frac{du}{dx} and when I do \frac{dy}{dx} I get \frac {(3x^2)(x^3-1)}{2(x^3)^\frac {3}{2}} is this correct? or is it just \frac {(3x^2)(x^3)}{2(x^3)^\frac {3}{2}}
    how is u related to x ?
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  3. #3
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    \frac {dy}{du} = \frac {1}{2}u^\frac{-1}{2}-\frac{1}{2}u^\frac{-3}{2}

    \frac{du}{dx} = 3x^2
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  4. #4
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    Quote Originally Posted by oregon88 View Post
    \frac {dy}{du} = \frac {1}{2}u^\frac{-1}{2}-\frac{1}{2}u^\frac{-3}{2}

    \frac{du}{dx} = 3x^2
    ok ... \frac{du}{dx} = 3x^2 ... is u = x^3 + 1000 ?
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  5. #5
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    u=x^3-x
    Last edited by oregon88; May 17th 2010 at 05:43 PM. Reason: Error
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  6. #6
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    \frac{d}{dx}\left[y = \sqrt{u} + \frac{1}{\sqrt{u}}\right]<br />

    \frac{dy}{dx} = \left[\frac{1}{2\sqrt{u}} - \frac{1}{2\sqrt{u^3}}\right] \cdot \frac{du}{dx}

    if u = x^3 - x

    \frac{du}{dx} = 3x^2 - 1

    \frac{dy}{dx} = \left[\frac{1}{2\sqrt{x^3-x}} - \frac{1}{2\sqrt{(x^3-x)^3}}\right]  \cdot (3x^2-1)
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  7. #7
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    When you simplify that you get \frac{(3x^2-1)(x^3-x-1)}{2(x^3-x)^\frac{3}{2}} but what if u=x^3? would you get \frac {(3x^2)(x^3-1)}{2(x^3)^\frac {3}{2}} or \frac {(3x^2)(x^3)}{2(x^3)^\frac {3}{2}}?
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