How Do You Solve for y' in the Derivative of sqrt(xy) + y^2 = 3x^3 * y^2?

[mbrod90]

find the derivative of ...

Homework Statement

find the derivative of
(sqrt(xy))+(y^2)=(3*x^3)*(y^2)
it has to equal to y'

The Attempt at a Solution

(-1/2(xy)^-1/2)(xy'+y)+(2yy')=(3x^3)(2yy')+(9x^2)(y^2)
i got stuck. idk how to get to y' ?

 

[Char. Limit]

Expand your equations first. Then get all terms containing y' on one side, and all terms not containing y' on the other side.

 

[mbrod90]

Expand your equations first. Then get all terms containing y' on one side, and all terms not containing y' on the other side.

i don't understand if all the terms have y in them how am i supposed to expand them? That is why I can't solve this.

 

[Char. Limit]

Well, expanding out all the equations, I get...

-(y' sqrt(x y))/(2 y)-sqrt(x y)/(2 x)+2 y y' = 6 x^3 y y'+9 x^2 y^2

Not all of those terms have y' in them. The terms that do, I put on one side of the equation. The terms that don't, I put on the other. Then you just factor y' out and then... the rest is obvious.

 

[mbrod90]

I don't understand what you did on the left side. I understand how to solve the problem from what you wrote though. thank you. can you please explain what you did?

 

[Char. Limit]

Well, first I expanded it to...

\frac{-xy'}{2\sqrt{xy}} - \frac{y}{2\sqrt{xy}} + 2yy'

Then I rationalized the denominators.

 

[mbrod90]

im stuck. sorry. i really don't know how u did it

 

[Char. Limit]

im stuck. sorry. i really don't know how u did it

Well, you know that c(a+b) = ac + ab, right? That's the principle I used here.

 

[mbrod90]

thats what i did. what did i do wrong.?

 

[mbrod90]

Well, you know that c(a+b) = ac + ab, right? That's the principle I used here.

you mean

a(c+b) ?

 

[Char. Limit]

you mean

a(c+b) ?

...yes. Sorry about that.

 

[mbrod90]

do you understand what i wrote? my handwriting is horrible. sorry.

 

[Char. Limit]

No, that looks right.

 

[mbrod90]

im sorry i fixed it. this is where i got stuck.

 

[Char. Limit]

The first thing at the bottom should be (xy' + y), not (xy' + y').

 

[HallsofIvy]

Homework Statement

find the derivative of
(sqrt(xy))+(y^2)=(3*x^3)*(y^2)
it has to equal to y'

The Attempt at a Solution

(-1/2(xy)^-1/2)(xy'+y)+(2yy')=(3x^3)(2yy')+(9x^2)(y^2)
i got stuck. idk how to get to y' ?

First, it is not "-1/2". The derivative of x^{1/2} is (1/2)x^{-1/2}
so this would be (1/2)(xy)^{-1/2}(xy'+ y)+ 2yy'= (9x^2)y^2+ 6x^3yy'
Now combine the terms involving y':
(1/2)(xy)^{-1/2}xy'+ 2yy'- 6x^3yy'= 9x^2y^2- (1/2)(xy)^{-1/2}y
[(1/2)(xy)^{-1/2}x+ 2y- 6x^3y]y'= 9x^2y^2- (1/2)(xy)^{-1/2}y
and solve for y'.