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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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Q: How do you prove that the derivative of sec x is equal to sec x tan x?
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