Question

find dy/dx: sin(mx) over x
I need suggestions
mcosx^2(mx)-(x-1)/(X^2)^2

NOT SO SURE

Answer 1

loos still wrong

Answer 2

*looks

Answer 3

consider sin(3x) * x^(-1) as a product rule

Answer 4

use the division rule, also the chain rule
f = u/v then f' = (vu' - uv')/v^2

Answer 5

hint :
y = sin(mx)/ x
u = sin(mx) ----> u' = mcos(mx)
v = x ----> v' = 1

now distribute all into formula above

Answer 6

I can see
The bottom one is wrong