If 3x^2+2xy+y^2=2 then what is the value of y'(1)?

My steps so far:

6x+2xy'+2y+2yy'=0

3x+xy'+y+yy'=0

Yy'+xy'=-3x-y

Y'=(-3x-y)/(y+x)

Would it be possible to find this or would the answer not be defined due to the two variables?

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If 3x^2+2xy+y^2=2 then what is the value of y'(1)?

My steps so far:

6x+2xy'+2y+2yy'=0

3x+xy'+y+yy'=0

Yy'+xy'=-3x-y

Y'=(-3x-y)/(y+x)

Would it be possible to find this or would the answer not be defined due to the two variables?

My steps so far:

6x+2xy'+2y+2yy'=0

3x+xy'+y+yy'=0

Yy'+xy'=-3x-y

Y'=(-3x-y)/(y+x)

Would it be possible to find this or would the answer not be defined due to the two variables?

- Coolmpl2
**Posts:**2**Joined:**Thu Oct 11, 2012 12:02 am

Coolmpl2 wrote:If 3x^2+2xy+y^2=2 then what is the value of y'(1)?

When x = 1, what is the correspondent value of y, based on the above equation? Use this to complete the computations.

- nona.m.nona
**Posts:**252**Joined:**Sun Dec 14, 2008 11:07 pm

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