Solve this problem to get a cookie:

if f(x) = [-cos(sin(3x^2))]/[tan(3x)]
find f''(x)

(Yes this problem is quite tedious, but if you want the cookie, start working)

f' = (sin(sin(3x^2))*cos(3x^2)*6x)/(tan(3x)) + (cos(sin(3x^2)))/(sin(3x)^2)

f'' = (cos(sin(3x^2))*cos(3x^2)^2*36x^2 - sin(sin(3x^2))*sin(3x^2)*36x^2 + sin(sin(3x^2))*cos(3x^2)*6)/(tan(3x)) + (sin(sin(3x^2))*cos(3x^2)*6x)/(sin(3x)^2) - (sin(sin(3x^2))*cos(3x^2)*6x)/(sin(3x)^2) - (cos(sin(3x^2))*cos(3x)*12)/(sin(3x))

I think.