Question

find the eqt. of the tangent to the curves at (1,2).
y^3+y^3=9(xy-1)
[ans: y=5x-5]
could some1 show me the calculation work bcus my ans is y=(6x+4)/5.

Answer 1

I will guess that you have a typo and one of those first terms is x^3

x^3 + y^3 = 9xy - 9
3x^2 + 3y^2 dy/dx = 9x dy/dx + 9y
dy/dx(3y^2 - 9x) = 9y - 3x^2

at (1,2)
dy/dx(12-9) = 18 - 3
dy/dx = 15/3 = 5

so equation is y = 5x + b
plug in (1,2)
2 = 5(1) + b
b = -3

tangent equation : y = 5x - 3

Their answer is wrong, since (1,2) doesn't even satisfy their answer equation.

Answer 2

the 1st term is y^3

Answer 3

Then your equation would simplify to
2y^3 = 9xy - 9

6y^2 dy/dx = 9x dy/dx + 9y
dy/dx(6y^2 - 9x) = 9y
for the given point (1,2)
dy/dx(24-9) = 18
dy/dx = 18/15 = 6/5

Now the slope of the line will be 6/5 , but in your answer it is 5, as I had before.
Now it makes even less sense!

Please check your typing, it is frustrating to keep guessing what the student meant in solving these questions.

Answer 4

you get the same ans like me so the ans is wrong.
Is that right?

Answer 5

It appears that way, but I still think no textbook would have an equation looking like

y^3 + y^3 .....