Find dy/dx of?

Question

x^3+3xy^2+y^3=1

Answer 1

Using implicit differentiation of

x^3 + 3xy^2 + y^3 = 1

We get

3x^2 + 3[y^2 + x(2y)(dy/dx)] + 3y^2(dy/dx) = 0

3x^2 + 3y^2 + 6xy(dy/dx) + 3y^2(dy/dx) = 0

Keeping everything with a (dy/dx) on the left hand side, and moving everything else to the right hand side, we have

6xy(dy/dx) + 3y^2(dy/dx) = -3x^2 - 3y^2

Factoring out dy/dx, we have

(dy/dx) [6xy + 3y^2] = -3x^2 - 3y^2

And then dividing both sides to isolate dy/dx, we have

dy/dx = (-3x^2 - 3y^2) / (6xy + 3y^2)

We can factor it slightly

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

dy/dx = (-1/y) (x^2 + y^2) / (2x + y)

EDIT: Thanks to Tomato Paste for pointing out my mistake. This is now edited for the correct answer.

The lesson here is to always do your math questions in steps, one at a time, because it makes the mistake pinpointable.

Answer 2

There is one right answer among these. The fast answer is correct. I don't answer these questions because It doesn't help you to give answers without an explanation. It needs to make sense before you can truly remember it and calculus takes extreme focus to understand. But I'll tell you that the "fast answer" is the correct answer. The others have either made slight errors or completely forgot steps.

Again the right answer is (-3x^2-3y^2)/(6xy+37^2) or as the person with the "fast answer" has it with the 3 divided out of the top and bottom and minus factored out as

-(x^2+y^2)/(2xy+y^2) which is fine also.

Answer 3

differentiate the problem. (Pull down the power, making it the coefficient, and decreasing the power by 1. Do this with y also, but multiply it by dy/dx)
3x^2 + 3*2y(dy/dx) + 3y^2(dy/dx) = 0

= 3x^2 + 6y(dy/dx) + 3y^2(dy/dx) = 0

Get the dy/dx terms singled out
6y(dy/dx) + 3y^2(dy/dx) = -3x^2

Factor out dy/dx
(dy/dx)(6y + 3y^2) = -3x^2

Get dy/dx alone
dy/dx = (-3x^2)/(6y + 3y^2)

factor out the common factors
dy/dx = 3(-x^2)/3(2y+y^2)

Simplify to obtain answer
dy/dx = -x^2/(2y + y^2)

Answer 4

Fast answer:

3x^2 + 6xy(dy/dx) + 3y^2 + 3y^2(dy/dx) = 0;

dy/dx(6xy + 3y^2) = -(3x^2 + 3y^2);

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

Answer 5

3x^2 + 3(2y)(dy/dx) + (3y^2)(dy/dx) = 0
dy/dx = -(3x^2)/(6y + 3y^2)