Equation of the Tangent Line
posted by Stewart .
Find the equation of a line tangent to te curve xy = sqrt(xy  x) + 1 at the point (1, 2).

xy = (xy  x)^(1/2)
x dy/dx + y = (1/2)(xy  x)^(1/2) (xdy/dx + y  1)
so at (1,2)
dy/dx + 2 = (1/2)(2  1)^(1/2) (dy/dx + 2  1)
times 2
2dy/dx + 4 = (1)(dy/dx + 1)
dy/dx = 14 = 3
so tangent is y = 3x + b
with (1,2) lying on it
2 = 3+b
b=5
tangent equation:
y = 3x + 5
check my arithmetic
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