Posted by **sara** on Saturday, October 20, 2007 at 5:54pm.

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

Find all points on the curve where x=2. Show there is a horizontal tangent to the curve at one of those points.

## Answer this Question

## Related Questions

- calculus - dy/dx= (2x-2xy)/(x^2+2y) find all points on the curve where x=2. show...
- calculus - dy/dx= (2x-2xy)/(x^2+2y) Find all points on the curve where x=2. Show...
- Calculus - Consider the curve y^2+xy+x^2=15. What is dy/dx? Find the two points ...
- calculus - x^2-x^2y=y^2-1 find all points on the curve where x=2. Show there is ...
- Calculus - Consider the curve given by y^2 = 2+xy (a) show that dy/dx= y/(2y-x...
- Calculus - the curve: (x)(y^2)-(x^3)(y)=6 (dy/dx)=(3(x^2)y-(y^2))/(2xy-(x^3)) a...
- Calculus - Can someone show how this question is solved. Consider the curve ...
- calculus - Given the curve defined by the equation y=cos^2(x) + sqrt(2)* sin(x) ...
- Calculus - Please help this is due tomorrow and I dont know how to Ive missed a ...
- Calculus - a)The curve with equation: 2y^3 + y^2 - y^5 = x^4 - 2x^3 + x^2 has ...

More Related Questions