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About LinkBacksIt goes through the point, so to evaluate y' at that point, substitute
and
. This will give you the slope of the tangent line. You know that it goes through that point, so you can find the equation of the tangent line.
The question is "Use implicit differentiation to write the equation of the tangent line to the graph at the specified point"
I'm given (x^2) - (2xy) + (y^2) - (2x) - (2y) + 1 = 0 with (2,3+2sqrt(2))
I tried to differentiate by the way I know getting 2x - (2xy' + 2y) + 2yy' - 2 - 2y' = 0 which simplies to (x-y-1)/(x-y+1) = y'
So I'm stuck here, is this it? am I done with the problem? I don't know why they gave me (2,3+2sqrt(2))
Can anyone help me please... I'm stuck..
Thank you
It goes through the point
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