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Math Help - what is the dy of a trigonometric function

  1. #1
    Member dokrbb's Avatar
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    what is the dy of a trigonometric function

    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
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  2. #2
    Member dokrbb's Avatar
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    Re: what is the dy of a trigonometric function

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

    please help,
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