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Old 07-30-2007   #2
Vipin A
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to arrive at ttangent to a curve at a specific point, differentiate the equation and plug in the point to find the slope of tangent at that point. then form the tangent equationdf(x)/dx = -2/x^3at (1,1), df(1)/dx = -2so tangent at (1,1) is y-1 = -2(x-1)y + 2x = 3