We have to find the derivative of f(x)=tan(2x)sin(x)
I would like to know why my approach is incorrect:
f′(x)=sin(x)⋅2cos2(2x)−tan(2x)⋅cos(x)sin2(x)
=2sin(x)−tan(2x)⋅cos(x)cos2(2x)⋅sin2(x)
=2sin(x)−sin(2x)⋅cos(x)cos3(2x)⋅sin2(x)
p.s. - To avoid confusion ; I wanted to get rid of the tan. I'm sure there is a shorter method than this but I don't want it; I just want to know why this is wrong.
Answer
Third line is: 2sin(x)−tan(2x)⋅cos(x)⋅cos2(2x)cos2(2x)⋅sin2(x)
instead of 2sin(x)−tan(2x)⋅cos(x)cos2(2x)⋅sin2(x)
No comments:
Post a Comment