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.

1. anonymous

You do that thing called implicit differentiation.

2. anonymous

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

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

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous