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Show that sec'x = d/dx (sec x) = sec x tan x. First, take note that sec x = 1/cos x; d sin x = cos x dx; d cos x = -sin x dx; and d log u = du/u. From the last, we have du = u d log u. Then, letting u = sec x, we have, d sec x = sec x d log sec x; and d log sec x = d log ( 1 / cos x ) = -d log cos x = d ( -cos x ) / cos x = sin x dx / cos x = tan x dx. Thence, d sec x = sec x tan x dx, and sec' x = sec x tan x, which is what we set out to show.
First change mi/hr into ft/sec. There 1 mi = 5280 ft and 1 hour = 3600 sec. 6 mi/hr = 6 * (5280 ft / 3600 sec) = 8.8 ft / sec To figure out how fast you travel any distance you divide the distance by time it took you. distance / time = speed 205 ft / time = 8.8 ft / sec Solving for time: time = 205 ft / (8.8 ft / sec) time = 23 sec
Write sec x as a function of sines and cosines (in this case, sec x = 1 / cos x). Then use the division formula to take the first derivative. Take the derivative of the first derivative to get the second derivative. Reminder: the derivative of sin x is cos x; the derivative of cos x is - sin x.
Sir Robert Walpole, the first Prime Minister, at that time First Lord of the Treasury, of the UK.