Answer
$$\frac{d}{{dx}}\left[ {\sec x} \right] = \sec x\tan x$$
Work Step by Step
$$\eqalign{
& \sec x = \frac{1}{{\cos x}} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\sec x} \right] = \frac{d}{{dx}}\left[ {\frac{1}{{\cos x}}} \right] \cr
& {\text{by using the quotient rule}} \cr
& \frac{d}{{dx}}\left[ {\sec x} \right] = \frac{{\left( {\cos x} \right)\frac{d}{{dx}}\left[ 1 \right] - \frac{d}{{dx}}\left[ {\cos x} \right]}}{{{{\left( {\cos x} \right)}^2}}} \cr
& {\text{then}} \cr
& \frac{d}{{dx}}\left[ {\sec x} \right] = \frac{{\left( {\cos x} \right)\left( 0 \right) - \left( { - \sin x} \right)}}{{{{\left( {\cos x} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{d}{{dx}}\left[ {\sec x} \right] = \frac{{\sin x}}{{{{\left( {\cos x} \right)}^2}}} \cr
& \frac{d}{{dx}}\left[ {\sec x} \right] = \left( {\frac{1}{{\cos x}}} \right)\left( {\frac{{\sin x}}{{\cos x}}} \right) \cr
& \frac{d}{{dx}}\left[ {\sec x} \right] = \sec x\tan x \cr} $$