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Math Help - Derivative Help

  1. #1
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    Derivative Help

    I am having a lot of trouble with the following problem:

    Recall that and that .
    The ratio is a constant number. Its value is....



    I keep getting: 8(cosx)/sin^3x




    Doesn't seem to be the right answer. Any help would be great! Thanks!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jwebb19 View Post
    I am having a lot of trouble with the following problem:

    Recall that and that .
    The ratio is a constant number. Its value is....



    I keep getting: 8(cosx)/sin^3x




    Doesn't seem to be the right answer. Any help would be great! Thanks!
    \frac{d}{dx}cot(x) = -(cot^2(x) + 1) = -csx^2(x)

    To understand how this is happening, see if you can derive the previous formulas I used.

    -Dan
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  3. #3
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    Thanks!


    Now a new problem.......

    I cannot get the derivative of



    I get (3.5+(3/(2sqrt(3x))))/sqrt(7x+sqrt(3x))
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jwebb19 View Post
    Thanks!


    Now a new problem.......

    I cannot get the derivative of



    I get (3.5+(3/(2sqrt(3x))))/sqrt(7x+sqrt(3x))
    Please post new questions in new threads.

    y = \sqrt{7x + \sqrt{3x}}

    Use the chain rule:
    \frac{dy}{dx} = \frac{1}{2\sqrt{7x + \sqrt{3x}}} \cdot \left ( 7 + \frac{1}{2\sqrt{3x}} \cdot 3 \right )

    You simplify it.

    -Dan
    Last edited by topsquark; October 4th 2007 at 05:34 AM. Reason: Doh!
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  5. #5
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    Quote Originally Posted by topsquark View Post
    Please post new questions in new threads.

    y = \sqrt{7x + \sqrt{3x}}

    Use the chain rule:
    \frac{dy}{dx} = \frac{1}{2\sqrt{7x + \sqrt{3x}}} \cdot \left ( 7 + \frac{1}{2\sqrt{3x}} \right ) \cdot 3

    You simplify it.

    -Dan

    \frac{dy}{dx} = \frac{1}{2\sqrt{7x + \sqrt{3x}}} \cdot \left ( 7 + \frac{\sqrt{3}}{2\sqrt{x}} \right )
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    \frac{dy}{dx} = \frac{1}{2\sqrt{7x + \sqrt{3x}}} \cdot \left ( 7 + \frac{\sqrt{3}}{2\sqrt{x}} \right )
    Thanks for the spot. I fixed it in the post.

    -Dan
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