Calculus
posted by Jeremy .
Find an equation of the tangent line to the curve at the given point
y^2 = x^3 (2x) (1,1)

Calculus 
Steve
y' = 3x^2 (2x)  x^3
y'(1) = 3(1)  1 = 2
so, now yu have a slope and a point:
y1 = 2(x1)
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