Answer
$$t'\left( x \right) = \frac{{ - {{\csc }^2}x - \sec x{{\csc }^2}x - \sec x}}{{{{\left( {1 + \sec x} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& t\left( x \right) = \frac{{\cot x}}{{1 + \sec x}} \cr
& {\text{Differentiate}} \cr
& t'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{\cot x}}{{1 + \sec x}}} \right] \cr
& {\text{By the quotient rule}} \cr
& t'\left( x \right) = \frac{{\left( {1 + \sec x} \right)\left( {\cot x} \right)' - \cot x\left( {1 + \sec x} \right)'}}{{{{\left( {1 + \sec x} \right)}^2}}} \cr
& {\text{Computing derivatives}} \cr
& t'\left( x \right) = \frac{{\left( {1 + \sec x} \right)\left( { - {{\csc }^2}x} \right) - \cot x\left( {\sec x\tan x} \right)}}{{{{\left( {1 + \sec x} \right)}^2}}} \cr
& {\text{Simplifying}} \cr
& t'\left( x \right) = \frac{{ - {{\csc }^2}x - \sec x{{\csc }^2}x - \sec x}}{{{{\left( {1 + \sec x} \right)}^2}}} \cr} $$
