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