derivative of f(x) = √( ( tan( csc(√x) ) ) / ( sin³x * cot(2x) ) )?

f = √[tan(csc√x)/(sin^3(x) cot(2x))]

df/dx = 1/2 * 1/√[tan(csc√x)/(sin^3x cot2x)] * d[tan(csc√x)/(sin^3x cot2x)]/dx
d[tan(csc√x)/(sin^3x cot2x)]/dx = dtan(csc√x)/dx * 1/(sin^3x cot2x) + d(sin^3x cot2x)^(-1)/dx * tan(csc√x)
= sec^2(csc√x) * d(csc√x)/dx * 1/(sin^3x cot2x) - 1/(sin^3x cot2x)^2 * d(sin^3x cot2x)/dx * tan(csc√x)
d(csc√x)/dx = -cot(√x)csc(√x) * d√x/dx
d√x/dx = 1/(2√x)
d(sin^3x cot2x)/dx = d(sin^3x)/dx * cot2x + d(cot2x)/dx * sin^3x
d(sin^3x)/dx = 3sin^2x * d(sinx)/dx
d(sinx)/dx = cosx
d(cot2x)/dx = -csc^2(2x) * d(2x)/dx
d(2x)/dx = 2

Therefore (substitute the components back into the df/dx),
d(cot2x)/dx = -2csc^2(x)
d(sin^3x)/dx = 3sin^2x * cosx
d(sin^3x cot2x)/dx = 3sin^2x * cosx * cot2x - 2csc^2(x) * sin^3x
d(csc√x)/dx = -1/(2√x) * cot(√x)csc(√x)
d[tan(csc√x)/(sin^3x cot2x)]/dx = -1/(2√x) * sec^2(csc√x) * cot(√x)csc(√x) * 1/(sin^3x cot2x) - 1/(sin^3x cot2x)^2 * (3sin^2x * cosx * cot2x - 2csc^2(x) * sin^3x) * tan(csc√x)
df/dx = 1/2 * 1/√[tan(csc√x)/(sin^3x cot2x)] * (-1/(2√x) * sec^2(csc√x) * cot(√x)csc(√x) * 1/(sin^3x cot2x) - 1/(sin^3x cot2x)^2 * (3sin^2x * cosx * cot2x - 2csc^2(x) * sin^3x) * tan(csc√x))

You may try to simplify...