Differentiate the function f(x) = 3x^2/sin(2x)

Using the product rule, f=uv, df = (vu'-uv')/v^2. we first set u = 3x^2 and v = sin(2x). u' = 6x, v'=2cos(2x) Therefore, vu' = 6xsin(2x). uv' = 6x^2cos(2x), v^2 = 4cos^2(2x) Therefore the differential is [6xsin(2x) - 6x^2cos(2x)]/[4cos^2(2x)] We can factor out 6x from the top and divide by the 4 on the bottom to give 3x(sin(2x)-xcos(2x))/(2*cos^2(2x))