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

Algebra ->  Test -> SOLUTION: Find the equation of the line tangent to f(x)=sin(2x)ln(x^2) at the point (1,0)      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!

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