Answer
$$\frac{{dy}}{{dx}} = \frac{{ - 1 - 6\tan x + x\tan x}}{{\sec x}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{6 - x}}{{\sec x}} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}\left[ {\frac{{6 - x}}{{\sec x}}} \right] \cr
& {\text{use the quotient rule for derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\sec x \cdot {D_x}\left( {6 - x} \right) - \left( {6 - x} \right) \cdot {D_x}\left( {\sec x} \right)}}{{{{\left( {\sec x} \right)}^2}}} \cr
& {\text{solving derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\sec x\left( { - 1} \right) - \left( {6 - x} \right)\left( {\sec x\tan x} \right)}}{{{{\left( {\sec x} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{ - \sec x - 6\sec x\tan x + x\sec x\tan x}}{{{{\sec }^2}x}} \cr
& \frac{{dy}}{{dx}} = \frac{{ - 1 - 6\tan x + x\tan x}}{{\sec x}} \cr} $$
