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Math Help - Implicit differentiation

  1. #1
    Junior Member
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    Implicit differentiation

    Hi, please help!

    question is

    Find all points on the curve x^2y^2+xy=2 where the slope of the tangent is –1.

    d/dx (2xy) + (2xy) dy/dx + (y) + (x) dy/dx = 0

    Why is " y + x (dy/dx) = 0 " , shouldn't it be " x + y (dy/dx) = 0 " Kindly explain.

    dy/dx (2xy + x) = -2x y-y
    dy/dx = -2x y-y / 2xy + x

    does it then = x - y / y + x
    and does that mean it = -1 because of " = x - y / y + x "

    Many thanks for you help if your can explain this!
    Last edited by Jon123; May 9th 2011 at 07:49 AM.
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  2. #2
    Senior Member Sambit's Avatar
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    Quote Originally Posted by Jon123 View Post
    Why is " y + x (dy/dx) = 0 " , shouldn't it be " x + y (dy/dx) = 0 " Kindly explain.
    I think you mean the differentiating the part xy w.r.t. x.

    According to the product rule of differentiation, \frac{d}{dx}(xy)=x\frac{d}{dx}(y)+y\frac{d}{dx}(x)  =x\frac{dy}{dx}+y\frac{dx}{dx}=x\frac{dy}{dx}+y. (first fix x and differentiate y w.r.t y; then fix y and differentiate x w.r.t y)

    dy/dx (2xy + x) = -2x y-y
    dy/dx = -2x y-y / 2xy + x
    You made mistake in this part.


    \frac{d}{dx}(x^2y^2+xy) = 2xy^2+2x^2y\frac{dy}{dx}+x\frac{dy}{dx}+y.

    So \frac{dy}{dx} = \frac{-y-2xy^2}{x+2x^2y}. This is your slope. Equate this with -1 and simplify to get the answer.
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