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Thread: Determining dy/du, du/dx, dy/dx

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

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

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

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

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

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

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

    if $\displaystyle u = x^3 - x$

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

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