Posted by **Anonymous** on Monday, August 5, 2013 at 2:14pm.

Find the equation of the tangent line to the curve x^(3)+3xy+y^(3)=1 at the point (1,0).

- Calculus -
**Reiny**, Monday, August 5, 2013 at 3:05pm
Implicit derivative:

3x^2 + 3x dy/dx + 3y + 3y^2 dy/dx = 0

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

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

at (1,0)

dy/dx = -1 - 0)/(1 + 0) = -1

let y = -x + b

at (1,0)

0 = -1 + b

b = 1

tangent equation at (1,0) is

y = -x + 1

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