Answer
$$\frac{{dy}}{{dx}} = - \frac{{{{\cot }^2}x{{\csc }^2}x}}{{\sqrt {1 + {{\cot }^2}x} }}$$
Work Step by Step
$$\eqalign{
& y = \sqrt {1 + {{\cot }^2}x} \cr
& {\text{Write as}} \cr
& y = {\left( {1 + {{\cot }^2}x} \right)^{1/2}} \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\left( {1 + {{\cot }^2}x} \right)}^{1/2}}} \right] \cr
& {\text{Use the chain rule}}{\text{,}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {1 + {{\cot }^2}x} \right)^{1/2 - 1}}\frac{d}{{dx}}\left[ {1 + {{\cot }^2}x} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {1 + {{\cot }^2}x} \right)^{ - 1/2}}\left( {2\cot x} \right)\frac{d}{{dx}}\left[ {\cot x} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{{{{\left( {1 + {{\cot }^2}x} \right)}^{1/2}}}}\left( {\cot x} \right)\left( { - {{\csc }^2}x\cot x} \right) \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = - {\left( {1 + {{\cot }^2}x} \right)^{ - 1/2}}\left( {{{\cot }^2}x{{\csc }^2}x} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{{{{\cot }^2}x{{\csc }^2}x}}{{\sqrt {1 + {{\cot }^2}x} }} \cr} $$
