Possible derivation:
d/dx(tan^(-1)(2 x) sin(2 x))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = tan^(-1)(2 x) and v = sin(2 x):
= tan^(-1)(2 x) (d/dx(sin(2 x)))+(d/dx(tan^(-1)(2 x))) sin(2 x)
Using the chain rule, d/dx(sin(2 x)) = ( dsin(u))/( du) 0, where u = 2 x and ( d)/( du)(sin(u)) = cos(u):
= (d/dx(tan^(-1)(2 x))) sin(2 x)+cos(2 x) d/dx(2 x) tan^(-1)(2 x)
Factor out constants:
= (d/dx(tan^(-1)(2 x))) sin(2 x)+2 d/dx(x) tan^(-1)(2 x) cos(2 x)
The derivative of x is 1:
= (d/dx(tan^(-1)(2 x))) sin(2 x)+1 2 tan^(-1)(2 x) cos(2 x)
Using the chain rule, d/dx(tan^(-1)(2 x)) = ( dtan^(-1)(u))/( du) 0, where u = 2 x and ( d)/( du)(tan^(-1)(u)) = 1/(1+u^2):
= 2 tan^(-1)(2 x) cos(2 x)+(d/dx(2 x))/(4 x^2+1) sin(2 x)
Factor out constants:
= 2 tan^(-1)(2 x) cos(2 x)+2 d/dx(x) (sin(2 x))/(1+4 x^2)
The derivative of x is 1:
Answer: | = 2 tan^(-1)(2 x) cos(2 x)+1 (2 sin(2 x))/(1+4 x^2)
