Correct Answer - Option 2 :

\(\frac{\ - \ cosec^2 x}{2\sqrt{\cot x}}\)
__Concept:__

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

- Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
- Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)

\(\rm \frac{d\cot x}{dx} = -\ cosec^2 x\)

__Calculation:__

Let y = \(\rm \sqrt{\cot x}\)

Differentiating with respect to x, we get

\(\rm \Rightarrow \frac{dy}{dx} = \frac{d\sqrt{\cot x}}{dx}\\=\frac{d\sqrt{\cot x}}{d \cot x} \times \frac{d\cot x}{dx}\\=\frac{1}{2\sqrt{\cot x}} \times(- cosec^2 x) \\\therefore \frac{dy}{dx} =\frac{\ -\ cosec^2 x}{2\sqrt{\cot x}}\)