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.

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.

the 1st term is y^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.

you get the same ans like me so the ans is wrong.

Is that right?

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

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

To find the equation of the tangent to the curve at the point (1,2), we need to first find the derivative of the curve. Then we can substitute the coordinates of the given point into the equation of the tangent line.

1. Start by finding the derivative of the equation of the curve.
The equation is given as y^3 + y^3 = 9(xy - 1).
To differentiate, we need to simplify the equation:
2y^3 = 9xy - 9.

2. Differentiate both sides of the equation with respect to x.
Differentiating 2y^3 with respect to x gives us:
6y^2 * dy/dx = 9y + 9x * dy/dx.

3. Rearrange and simplify the equation.
Multiplying through by dx, we have:
6y^2 * dy = (9y + 9x) * dx.

4. Divide both sides of the equation by 6y^2.
dy/dx = (9y + 9x) / (6y^2).

5. Now substitute the coordinates of the given point (1,2) into the equation.
dy/dx = (9(2) + 9(1)) / (6(2)^2) = (18 + 9) / (6 * 4) = 27 / 24 = 9 / 8.

6. The slope of the tangent line is dy/dx, so it is 9/8.

7. Use the point-slope form of a line to find the equation of the tangent line.
Using the point (1,2) and the slope 9/8, we have:
y - 2 = (9/8)(x - 1).

8. Simplify the equation.
y - 2 = 9x/8 - 9/8.
Multiply through by 8 to eliminate the fraction:
8(y - 2) = 9x - 9.
8y - 16 = 9x - 9.
Rearrange the equation to get the standard form:
9x - 8y + 7 = 0.

Therefore, the equation of the tangent line to the curve at point (1,2) is 9x - 8y + 7 = 0, which is equivalent to y = (9/8)x - 7/8.

It seems like your answer, y = (6x + 4)/5, is incorrect. Please double-check your calculations.