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

## Answer This Question

## Related Questions

- Calculus - Please help this is due tomorrow and I dont know how to Ive missed a ...
- calculus - what is the equation of the tangent line to the curve x^3 + 2y^2 + ...
- calculus - Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 . a. Show ...
- Calculus - If F(x)=x^3−7x+5, use the limit definition of the derivative to...
- Calculus AP - implicit differentiation Find the slope of the tangent line to ...
- Calculus AP need help! - i dont know how or what to starts Find the slope of the...
- Calculus - 2y^3 - 3xy = 4 1) Find dy/dx 2) Write an equation for the line ...
- AP AB Calculus - Linear approximation: Consider the curve defined by -8x^2 + 5xy...
- calculus - Consider line segments which are tangent to a point on the right half...
- AP Calculus - Consider the curve given by x^2+4y^2=7+3xy a) Show that dy/dx=(3y-...

More Related Questions