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Math Help - implicit differenciation

  1. #1
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    implicit differenciation

    find y" of x+xy+y=2

    so i did the first differncition as

    1+y+x(dy/dx)+1(dy/dx) = 0
    y= 1+y/x+1

    y"= ?

    somebody solve this for me ?

    y" = ( 1+y)-(x+1)/ (x+1)^2

    i think this is not right .
    what is the answear. show me with working !
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  2. #2
    Senior Member BAdhi's Avatar
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    Re: implicit differenciation

    So you've already taken the first differentiation which is,

    1+y+x\left[ \frac{dy}{dx}\right] +\frac{dy}{dx}=0\qquad \qquad (1)

    from the given equation we know that,

    y=\frac{(2-x)}{(x+1)}

    for later use, let's isolate \frac{dy}{dx} from (1),

    \frac{dy}{dx}=\frac{-(1+y)}{(x+1)}=\frac{-\left( 1+\frac{(2-x)}{(x+1)}\right) }{(x+1)}=\frac{-3}{(x+1)^2}\qquad \qquad (2)


    to find y'', differentiate the equation (2) w.r.t x,
    Last edited by BAdhi; August 22nd 2012 at 11:43 PM. Reason: clarification of the expalnation
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  3. #3
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    Re: implicit differenciation

    why is ot

    -(1+y) ?

    i differentiated
    -3/(x+1)^2

    to get
    6/(x+1)^3

    what now . i dont get it
    Last edited by arsenal12345; August 22nd 2012 at 10:27 PM.
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  4. #4
    Senior Member BAdhi's Avatar
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    Re: implicit differenciation

    Sorry, I think that I was not much clear on the previous post. So I did some editing. Do you understand it now?
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  5. #5
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    Re: implicit differenciation

    6/(x+1)^3

    so the answear is that ?
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  6. #6
    Senior Member BAdhi's Avatar
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    Re: implicit differenciation

    Yes, I think it is
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  7. #7
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    Re: implicit differenciation

    Hello, arsenal12345!

    \text{Differentiate implicitly. \;Find }y"\text{of }x+xy+y\:=\:2

    Differentiate: . 1 + y + xy' + y' \:=\:0

    n . . . . . . . . . . . . . . (x+1)y' \:=\:-(y+1)

    . . . . . . . . . . . . . . . . . . . . y' \:=\:-\frac{y+1}{x+1} .[1]


    Differentiate: . . y" \:=\:-\frac{(x+1)y' - (y+1)1}{(x+1)^2}

    Substitute [1]: . y" \:=\:-\frac{(x+1)\left(-\frac{y+1}{x+1}\right) - (y+1)}{(x+1)^2}

    . . . . . . . . . . . . y" \:=\:\frac{y+1 + y+1}{(x+1)^2} \:=\:\frac{2(y+1)}{x+1)^2}
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