SOLUTION: Find the equation of the line tangent to f(x)=sin(2x)ln(x^2) at the point (1,0)

SOLUTION: Find the equation of the line tangent to f(x)=sin(2x)ln(x^2) at the point (1,0)

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Question 406561: Find the equation of the line tangent to f(x)=sin(2x)ln(x^2) at the point (1,0)
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!

Hence the equation of the tangent line is y = 2sin2(x-1)
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