Question

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.

Answer 1

You do that thing called implicit differentiation.

Answer 2

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'

Answer 3

so 3x^2 + 3y^2 dy/dx - 2 ( x dy/dx + y ) =0

Answer 4

\[3x^2 + 3y^2 \frac{dy}{dx} -2 ( x \frac{dy}{dx}+y ) =0 \]

Answer 5

so dy/dx { 3y^2 -2x} = -3x^2+2y\[\frac{dy}{dx} = \frac{-3x^2 +2y}{3y^2 -2x} \]

Answer 6

at (1,1) , dy/dx = -1\[y-y1=m(x-x1)\]\[(y-1)=-(x-1) \]\[y= -x+2 \]

Answer 7

^ answer

Answer 8

sorry for slow response but you may have notices later that it was a double post but thanks a ton