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anonymous

  • 5 years ago

Find the equation of the tangent line at the point (1,1) for the curve x^3+y^3-2xy=0. But I can't for the life of me figure out how to solve the equation for y.

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  1. anonymous
    • 5 years ago
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    You do that thing called implicit differentiation.

  2. anonymous
    • 5 years ago
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    differentiate both sides with respect to x , not for y^3 we need to apply chain rules and for xy we need to apply product rules'

  3. anonymous
    • 5 years ago
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    so 3x^2 + 3y^2 dy/dx - 2 ( x dy/dx + y ) =0

  4. anonymous
    • 5 years ago
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    \[3x^2 + 3y^2 \frac{dy}{dx} -2 ( x \frac{dy}{dx}+y ) =0 \]

  5. anonymous
    • 5 years ago
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    so dy/dx { 3y^2 -2x} = -3x^2+2y\[\frac{dy}{dx} = \frac{-3x^2 +2y}{3y^2 -2x} \]

  6. anonymous
    • 5 years ago
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    at (1,1) , dy/dx = -1\[y-y1=m(x-x1)\]\[(y-1)=-(x-1) \]\[y= -x+2 \]

  7. anonymous
    • 5 years ago
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    ^ answer

  8. anonymous
    • 5 years ago
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    sorry for slow response but you may have notices later that it was a double post but thanks a ton

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