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Find an equation of the tangent line to the graph y at the point x. y = 3x/(2x + x^-1) at x=1
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First you find the derivative of the function: y'=6x/(2x^2+1)^2 Check the value at x=1 y'(1)=2/3 the function has value at x=1 y(1)=1 So slope is 2/3 and takes the point 1,1 y=2/3x+1/3
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