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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.
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