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Find the equation of the line tangent to the curve x³ + y³ = 2xy at (1,1).
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[tex]y = -x +2[/tex]

Step-by-step explanation:

[tex]\text{Given that,}\\\\~~~~~~~~~~x^3 +y^3 = 2xy\\\\\implies \dfrac{d}{dx}(x^3 +y^3) = 2\dfrac{d}{dx}(xy)\\\\\implies 3x^2 +3y^2 \dfrac{dy}{dx} = 2 \left( y + x\dfrac{dy}{dx}\right)\\\\\implies 3x^2 + 3y^2 \dfrac{dy}{dx} = 2y + 2x\dfrac{dy}{dx}\\\\\implies 3y^2 \dfrac{dy}{dx}-2x \dfrac{dy}{dx} = 2y - 3x^2\\\\\implies (3y^2 -2x) \dfrac{dy}{dx} = 2y-3x^2\\\\\implies \dfrac{dy}{dx} = \dfrac{2y-3x^2}{3y^2 -2x}\\[/tex]

[tex]\text{At (1,1),}\\\\\text{Slope,}~ m = \dfrac{dy}{dx}\\\\~~~~~~~~~~~~~=\dfrac{2(1) -3(1^2)}{3(1^2) -2(1)}\\\\~~~~~~~~~~~~~=\dfrac{2-3}{3-2}\\\\~~~~~~~~~~~~~=\dfrac{-1}{1}\\\\~~~~~~~~~~~~~=-1[/tex]

[tex]\text{Equation of tangent line,}\\\\~~~~~~~~~y -y_1 = m(x-x_1)\\\\\implies y -1 = -1(x-1)\\\\\implies y -1 = -x+1\\\\\implies y = -x +1+1\\\\\implies y = -x +2[/tex]
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