Finding derivative and tangent

[mcale]

Homework Statement

Consider the curve given by xy^2 - x^3y = 6
a. Find 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.
c. Find the x-coordinate of each point on the curve where the tangent line is vertical/

Homework Equations

xy^2 - xy^3 = 6

The Attempt at a Solution

so i started with a and found the derivative to be y' = (3x^2*y - y^2)/(2xy - x^3) which was correct so no problems there

then i started b by plugging 1 for x into the original equation to get my y values which came out to be 3 and -2

then i plugged then into my equation for the derivative separately and found the derivative for 3 to be undefined and the derivative for -2 to be 2
i figured therefore there couldn't be any tangent equation for the y value of 3 so i used 2 and got the tangent to be y = 2(x-1)+2

im not sure everything after the derivative is correct
for part c i know that the equation for the tangent line has to just be straight line where it is just x = to a number but i don't know how to find that

 

[ehild]

You did a good job! The derivative is fine, and the points where x=1 are correct, but you made a small mistake in the equation of the tangent line in case of y=-2 (you used y=2 instead of y=-2).

then i plugged then into my equation for the derivative separately and found the derivative for 3 to be undefined

The derivative is not undefined. What do you get when you plug in x=1 and y=3 into the formula for y'?

for part c i know that the equation for the tangent line has to just be straight line where it is just x = to a number but i don't know how to find that

The derivative of the y(x) function is undefined when if it is a vertical line.
What is the denominator when y' is undefined?

You can also think of the inverse function x(y). If the tangent line of y(x) is vertical, that of x(y) is horizontal, that is dx/dy=0. It is the reciprocal of y' and it has to be zero.

ehild