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Math Help - chain rule and derivatives

  1. #1
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    Post chain rule and derivatives

    Compute the derivative dy/dx and simplify your answer.

    y = u^2 + u - 2
    u = 1/x or x^1/2

    y' = 2u + 1
    u' = 1/2(x)^-1/2 = 1/2(square root x)

    2(1/x) + 1 *1/2(square root x)
    2/x + 1 * 1/2(square root x)
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  2. #2
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    Quote Originally Posted by startingover
    Compute the derivative dy/dx and simplify your answer.

    y = u^2 + u - 2
    u= x^1/2
    y'=2u+1
    u=1/(2\sqrt{x})
    Thus,
    2(1/2(\sqrt{x}))+1=1/\sqrt{x}+1
    Multiply,
    1/(2\sqrt{x})(1/\sqrt{x}+1)
    Thus,
    1/(2x)+1/(2\sqrt{x})
    Last edited by ThePerfectHacker; August 26th 2006 at 08:12 PM.
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  3. #3
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    Unhappy Hmmmm....

    I'm sorry, but I'm a bit confused.

    I thought you substituted the orginal u expression (1/x) into the derivative of y (2u + 1), then multiplied by the derivative of u (1/2squ.rt.x).

    2[1/x] + 1 (1/2squ.rt.x)
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  4. #4
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    Quote Originally Posted by startingover
    I'm sorry, but I'm a bit confused.

    I thought you substituted the orginal u expression (1/x) into the derivative of y (2u + 1), then multiplied by the derivative of u (1/2squ.rt.x).

    2[1/x] + 1 (1/2squ.rt.x)
    Seems you know this stuff better than I do. Previous post has been corrected.
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  5. #5
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    Question ummmm...

    Not trying to be a pain - honest...

    In the beginning equation, why did you substitute 1/2square rt x in for the u expression? Shouldn't it be the orginal u, not the derivative of u?

    y = u^2 + u - 2
    y' = 2u + 1

    u = 1/x
    u' = 1/2squarert x

    2 (1/x) + 1 * 1/2squarert x
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  6. #6
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    Quote Originally Posted by startingover
    Compute the derivative dy/dx and simplify your answer.

    y = u^2 + u - 2
    u = 1/x or x^1/2

    y' = 2u + 1
    u' = 1/2(x)^-1/2 = 1/2(square root x)

    2(1/x) + 1 *1/2(square root x)
    2/x + 1 * 1/2(square root x)
    If u = \frac{1}{x} then u \ne x^{\frac{1}{2}}

    u = x^{-1}

    u' = -x^{-2}

    Have \frac{dy}{du} = 2u + 1 and \frac{du}{dx} = -x^{-2}

    \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

    = (2u + 1)(-x^{-2})

    = (2x^{-1} + 1)(-x^{-2})

    = -2x^{-3} - x^{-2}

    = -\frac{2}{x^3} -\frac{1}{x^2} = -\frac{2 + x}{x^3}
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  7. #7
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    Quote Originally Posted by startingover
    Compute the derivative dy/dx and simplify your answer.

    y = u^2 + u - 2
    u = 1/x or x^1/2

    y' = 2u + 1
    u' = 1/2(x)^-1/2 = 1/2(square root x)

    2(1/x) + 1 *1/2(square root x)
    2/x + 1 * 1/2(square root x)
    Using the ' notation for derivative here is confusing as it means d/dx in one
    place and d/dy in another.

    The chain rule is:

    dy/dx=dy/du du/dx

    so:

    dy/du=2u+1

    and:

    du/dx=-1/x^2 (that is when u=1/x)

    So:

    dy/dx=-(2u+1)/x^2=-(2/x+1)/x^2=-(x+2)/x^3

    RonL
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  8. #8
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    Quote Originally Posted by CaptainBlack
    Using the ' notation for derivative here is confusing as it means d/dx in one
    place and d/dy in another.

    The chain rule is:

    dy/dx=dy/du du/dx

    so:

    dy/du=2u+1

    and:

    du/dx=-1/x^2 (that is when u=1/x)

    So:

    dy/dx=(2u+1)/x^2=(2/x+1)/x^2=(x+2)/x^3

    RonL
    You missed out the minus from du/dx
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by Glaysher
    You missed out the minus from du/dx
    Noticed it and went back and put it in before seeing your note

    (peculiar that post does not show time it was edited)

    RonL
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