I checked and the results are wrong - I'm sure I'm doing some mistakes when applying the trig identities,
please help,
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