Sorry for posting the same thread, but the last one I described in such a confusing way even for me (no wonder I had no reply),

so, starting with a clean sheet:

The function y = tan(3x + 6); what is the dy when x=3 and dx=0.03

I tried these steps:

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

so, for f'(3) = (1/cos^2(3x+6))

= (1/cos^2(3(3)+6))

= (1/cos^2(15))

f'(3) = (1/cos^2(3x+6))

= (1/cos^2(3(3)+6))

= (1/cos^2(15))

= (1/[(1+cos2(15)/2)])

= (2/(1+cos30))

= {2/[1+ sqrt(3)/2]}

= [2/[(2+sqrt(3))/2]]

= 4/((2+sqrt(3)),

so, dy = (4/((2+sqrt(3)))(3)(0.03) = 0.36/(2+sqrt(3)) = and would be ..... am I doing this right