Derivative of Secant

Let $y = f(x) = \sec x$.

Since $\sec x = {1 \over {\cos x}} = (\cos x)^{ - 1} $, we use the chain rule and get

${{dy} \over {dx}} = ( - 1)(\cos x)^{ - 2} ( - \sin x)$

$ = {1 \over {\cos ^2 x}}\sin x$

$ = {1 \over {\cos x}}.{{\sin x} \over {\cos x}}$

$ = \sec x\tan x$