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Use implicit differentiation to find an equation of the tangent line to the curve x^2+xy+y^2=3 at the point (1,1)
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=2x+xy'+y+2yy'=0 =y'(x+2y)=2x+y y'=(2x+y)/(x+2y) plugging in (1,1) i get y'=1 plugging 1 into point-slope form i get y=x help?
2x+xy'+y+2yy'=0 gives y' (x+2y) = -2x-y isn't it ?
It sure does! Time for my coffee I guess, thanks :)
watch your signs!
need any further assistance with this??
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No, i've got this thanks!
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