I have the function y = tan(3x + 6); need to find dy when x=3 and dx=0.03

By the formula dy = f'(x)*dx ;

f'(x)= sec^2(3x+6)(4)dx = (1/cos^2(3x+6))(4)dx, (need to calculate in radians)

so,forf'(3) = (1/cos^2(3x+6)) = (1/cos^2(3(3)+6)) = (1/cos^2(15)) = (1/cos^2(45-30)).... am I on the right track , i'm confused on which identity to apply further

Just give me a hint, I don't need the solution,

Thank you