Find dy/dx by implicit differentiation: 6x^2+8xy+y^2=6

1. Jun 27, 2008

staples82

1. The problem statement, all variables and given/known data
Find dy/dx by implicit differentiation: 6x^2+8xy+y^2=6

2. Relevant equations
n/a

3. The attempt at a solution
I'm using y'=(dy/dx)

I found the derivative of the above problem.
12x+8xy'+10y=0 (I used the product rule to find the derivative of 8xy')

12x+8xy'+10y=0

8xy'=-12x-10y

xy'=(-12x-10y)/(8x)

I checked the answer in the back of the book and it was y'=(-6x-4y)/(4x+y)

I'm not sure where I missed a step or a calculation error

Thanks!!

2. Jun 27, 2008

rocomath

You forgot the chain rule on your y^2 term.

3. Jun 27, 2008

rootX

6x^2+8xy+y^2=6

12x+8xy' +8y + 2y**=0

This is wrong ...

4. Jun 27, 2008

jaredmt

i see what you tried to do. u tried to combine the 8y and 2y(dy/dx). they cannot be added together.

p.s.
next time u get stuck on something like this u can always work backwards. you could take the book's answer and multiply the top and bottom by 2. then u can figure it out from there

5. Jun 27, 2008

staples82

First of, thanks for all the help on my previous question, I have another one!

1. The problem statement, all variables and given/known data
Find dy/dx by implicit differentiation: 3x^2=(2-y)/(2+y)

2. Relevant equations
n/a

3. The attempt at a solution
I'm using y'=(dy/dx)

I found the derivative of the above problem.
First I used the quotient rule...
6x=[(2+y)(y')-(2-y)(y')]/(2+y)^2

6x=(2y'+yy'-2y'+yy')

6x=2yy'/(2+yy')^2

After this, I got into a jumbled mess, I wasn't sure if I had to add y' to the bottom of the quotient rule.

I checked the answer in the back of the book and it was y'=-3x(2+y)^2/2

Thanks!!

6. Jun 27, 2008

dynamicsolo

Don't forget that the first term in the numerator involves the derivative of (2 - y), so the quotient should be

[(2+y)(-y')-(2-y)(y')]/(2+y)^2 .

You could multiply by (2+y)^2 to get

6x · [(2+y)^2] = -2y' - yy' - 2y' + yy' = -4y'

and solve for y' .

7. Jun 27, 2008

Redbelly98

Staff Emeritus
Sign error: what's the derivitive of (2-y) ?

What happened to the denominator on the RHS?

How did (2+y)^2 from before become (2+yy')^2 ?