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Math Help - equations help ( differentiation )

  1. #1
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    equations help ( differentiation )

    The curve  y= cx + \frac{d}{x} has gradient 6 at the point  ( \frac{1}{2} , 1 ) find c and d

    my working,

    sub in the values for x and y and solve simultaneously but i dont get the right answer.....

    first equation =  \frac{1}{2}c + \frac{d}{\frac{1}{2}} which I simplfied to  2c + 2d = 1 is this simplification correct?

     \frac{dy}{dx} = c -dx^{-2}= 6

    these are my two equations, and when I try to solve for 'c' and 'd' i dont get the right answer?
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  2. #2
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    Hello,
    Quote Originally Posted by Tweety View Post
    The curve  y= cx + \frac{d}{x} has gradient 6 at the point  ( \frac{1}{2} , 1 ) find c and d

    my working,

    sub in the values for x and y and solve simultaneously but i dont get the right answer.....

    first equation =  \frac{1}{2}c + \frac{d}{\frac{1}{2}} which I simplfied to  2c + 2d = 1 is this simplification correct?
    It's 1=\frac 12 c +\frac{d}{\frac 12}
    There's a problem with your simplification :
    \frac{d}{\frac 12}=2d
    so you have 1=\frac 12 c+2d

     \frac{dy}{dx} = c -dx^{-2}= 6

    these are my two equations, and when I try to solve for 'c' and 'd' i dont get the right answer?
    This one is correct, though it is an equation for x=1/2
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  3. #3
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    how do i evaluate  -dx^{-2} when x =  \frac{1}{2}

    is this method correct..

     \frac{1}{2}^{-2} =  -\frac{1}{2} \times -\frac{1}{2} = \frac{1}{4}

    so the equation is  c -\frac{1}{4}d = 6 ?
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  4. #4
    Moo
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    Quote Originally Posted by Tweety View Post
    how do i evaluate  -dx^{-2} when x =  \frac{1}{2}

    is this method correct..

     \frac{1}{2}^{-2} =  -\frac{1}{2} \times -\frac{1}{2} = \frac{1}{4}

    so the equation is  c -\frac{1}{4}d = 6 ?
    wow... I think you've got some problems in algebra :s

    -x^{-2}=-\frac{1}{x^2}

    So when x=1/2, we have x^2=\left(\frac 12\right)^2=\frac{1^2}{2^2}=\frac 14

    ---> \frac{1}{x^2}=\frac{1}{\frac 14}=4

    -----> -dx^{-2}=-4d
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  5. #5
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    Quote Originally Posted by Moo View Post
    wow... I think you've got some problems in algebra :s

    -x^{-2}=-\frac{1}{x^2}

    So when x=1/2, we have x^2=\left(\frac 12\right)^2=\frac{1^2}{2^2}=\frac 14

    ---> \frac{1}{x^2}=\frac{1}{\frac 14}=4

    -----> -dx^{-2}=-4d
    Yes thats why I ask for help !!

    thanks anyways..........
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