2x + y + xy' + 2yy' = 0
You know x,y so just solve for y'.
Use implicit differentiation to find an equation of the tangent line to the curve at the given point x^2 + xy + y^2 = 3, (1,1)
2 answers
differentiate x^2+xy+y^2
f'= 2x+y+xy'+2yy'
f' =2(1)+(1)+(1)y'+2(1)y'
f'= 3+3y'
-3=3y'
y'=-1
to find tangent line,
y-y1=m(x-x1)
y-1= -1(x-1)
y-1= -x+1
x+y= 1+1
x+y=2
answer: x+y=2
f'= 2x+y+xy'+2yy'
f' =2(1)+(1)+(1)y'+2(1)y'
f'= 3+3y'
-3=3y'
y'=-1
to find tangent line,
y-y1=m(x-x1)
y-1= -1(x-1)
y-1= -x+1
x+y= 1+1
x+y=2
answer: x+y=2