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Math Help - Derivatives/Relative Rates

  1. #1
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    Derivatives/Relative Rates(unsolved)

    Hello, I have a couple questions, i'm not sure what im doing wrong.

    1.) find the derivative
    (x+1)^2(x^2+1)^-3
    so I do the product rule and the chain rule to get
    (2x)(x+1)^2-3(x^2+1)^-4+(x+1)(x^2+1)^-3
    is this correct so far? If yes, where do I go from here?
    I already have the answer I just cant figure out how to get there.

    2.) Evaluate dy/dt for the function at the point.
    x+y/x-y = x^2 + Y^2 , dx/dt = 12 , x = 1, y = 0

    I use the quotient rule on the left side and find the derivative or the right and get
    2 dy/dt = 2 dx/dt + 2 dy/dt
    the answer is 12, but I dont understand how you get there if you remove the dy/dt's. 2dy/dt - 2 dy/dt to take them to one side?? I can see that if you divide all 3 by 2 the dx/dt is 12 but there's no dy/dt's left in the problem for it to be equal to? do you see what im trying to say?


    Thanks
    Last edited by cokeclassic; May 11th 2010 at 01:14 PM.
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  2. #2
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    Quote Originally Posted by cokeclassic View Post
    1.) find the derivative
    (x+1)^2(x^2+1)^-3
    \frac{d}{dx}((x+1)^2(x^2+1)^{-3})= (x+1)^2((x^2+1)^{-3})'+((x+1)^2)'(x^2+1)^{-3}

    Now find ((x^2+1)^{-3})' and ((x+1)^2)'
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  3. #3
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    Quote Originally Posted by pickslides View Post
    \frac{d}{dx}((x+1)^2(x^2+1)^{-3})= (x+1)^2((x^2+1)^{-3})'+((x+1)^2)'(x^2+1)^{-3}

    Now find ((x^2+1)^{-3})' and ((x+1)^2)'
    ((x^2+1)^{-3})' and ((x+1)^2)'

    (x+1)^2-3(x^2+1)^-4(2x) + 2x= 2(x+1)(1)(x^2+1)^-3

    There we go is that correct? Where do I go after this step?
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  4. #4
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    Still need help with this


    edit: I think I understand what I did wrong on the second one, however i'm getting -12 for the answer and it's suppose to be 12.
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  5. #5
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    anyone?
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  6. #6
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    I can recall somewhere in the forum rules a part about no begging.

    ((x^2+1)^{-3})'  = -6x(x^2+1)^{-4}

    ((x+1)^2)' = (x^2+2x+1)' = 2x+2
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  7. #7
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    I understand what the derivative is.... i just don't understand how to simplify it
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  8. #8
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     \frac{dy}{dx} = \frac{-6x(x+1)^2}{(x^2+1)^{4}}+\frac{2x+2}{(x^2+1)^{3}}

     \frac{dy}{dx} = \frac{-6x(x+1)^2+(2x+2)(x^2+1)}{(x^2+1)^{4}}

    Expand then simplify the numerator by grouping like terms if you wish.
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