1. Consider the curve given by xy^2 - x^3 y =6
a) Show that dy/dx = 3x^2 y -y^2 / 2xy - x^3 (okay, I found the derivative)

b) Find all points on the curve whose x-coordinate is 1, and write an equation for the tangent line at each of these points.

(I'm not sure if I did this one correct.)
I plugged in 1 for the x's in the "original" equation: xy^2-x^3y =6
and I got y= -2 and 3
so my points were (1, -2) and (1, 3)

My equations are: (1,-2); y=2x-4
and for pt. (1,3); y=3


c) Find the x-coordinate of each point on the curve where the tangent line is vertical.
I'm not sure how to do this one.
I got this far: I set the denominator equal to 0;
(2xy-x^3)=0
2y=x^2


2. Given: x^2 - y^2 =16
a) Find dy/dx. (Found the derivative to be: -x/y)

b) Find the equation of the tangent line at (-5,3)
This is what I got for the equation, but I'm not sure if it's correct.
y=5/3x + 34/3

c) Find the second derivative.
I got: -y^2 - x^2 / y^

d) Determine the point(s) on x^2 - y^2 =16 where there is a vertical tangent.

e) Determine the point(s) on x^2 - y^2 =16 where there is a horizontal tangent.