Consider the equation x^2y+3y^3=24

Question

Find dy/dx

I got -2xy/x^2+9y^2

The next question is
Determine the points, if any where the tangent line to the graph of the equation is horizontal

How do I do this part??

Answer 1

9x-3y=24 , 11x+2y=1. 9x−3y=24 9 x - 3 y = 24 , 11x+2y=1 11 x + 2 y = 1

Answer 2

How did you get 9x or like 11x

Answer 3

x^2y+3y^3=24
x^2 dy/dx + y(2x) + 9y^2 dy/dx = 0
dy/dx(x^2 + 9y^2) = -2xy
dy/dx = -2xy/(x^2 + 9y^2)
your derivative would be correct if you included the brackets.

for the tangent to be horizontal it must have a zero slope, so
-2xy = 0
x = 0 or y = 0
if x = 0, 0 + 3y^3 = 24
y^3 = 8
y = 2, so at the point (0,2) the tangent is horizontal

if y = 0 , 0+0 = 24, which is not possible, so no point with y = 0

Answer 4

Thank you!