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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^32xy=0. But I can't for the life of me figure out how to solve the equation for y.
anonymous
 5 years ago
Find the equation of the tangent line at the point (1,1) for the curve x^3+y^32xy=0. But I can't for the life of me figure out how to solve the equation for y.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You do that thing called implicit differentiation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0differentiate 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'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 3x^2 + 3y^2 dy/dx  2 ( x dy/dx + y ) =0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[3x^2 + 3y^2 \frac{dy}{dx} 2 ( x \frac{dy}{dx}+y ) =0 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so dy/dx { 3y^2 2x} = 3x^2+2y\[\frac{dy}{dx} = \frac{3x^2 +2y}{3y^2 2x} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0at (1,1) , dy/dx = 1\[yy1=m(xx1)\]\[(y1)=(x1) \]\[y= x+2 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry for slow response but you may have notices later that it was a double post but thanks a ton
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