Answer
$$\frac{{dy}}{{dx}} = - \frac{1}{{2\sqrt {{{\cot }^{ - 1}}x} \left( {1 + {x^2}} \right)}}$$
Work Step by Step
$$\eqalign{
& y = \sqrt {{{\cot }^{ - 1}}x} \cr
& {\text{Write the function as}} \cr
& y = {\left( {{{\cot }^{ - 1}}x} \right)^{1/2}} \cr
& {\text{Differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{{\left( {{{\cot }^{ - 1}}x} \right)}^{1/2}}} \right) \cr
& {\text{Use the chain rule}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {{{\cot }^{ - 1}}x} \right)^{ - 1/2}}\frac{d}{{dx}}\left( {{{\cot }^{ - 1}}x} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {{{\cot }^{ - 1}}x} \right)^{ - 1/2}}\left( { - \frac{1}{{1 + {x^2}}}} \right) \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{2{{\left( {{{\cot }^{ - 1}}x} \right)}^{1/2}}}}\left( { - \frac{1}{{1 + {x^2}}}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{2\sqrt {{{\cot }^{ - 1}}x} \left( {1 + {x^2}} \right)}} \cr} $$
