Differentiate the functions with respect to x: $\;2\sqrt {cot (x^2)}$

$\begin{array}{1 1} \large \frac{-2\;x\;cosec^2x^2}{\sqrt cot x^2} \\ \large \frac{2\;x\;cosec^2x}{\sqrt cot x^2} \\ \large \frac{-2\;x\;sec^2x^2}{\sqrt cot x^2} \\ \large \frac{2\;x\;cosec^2x^2}{\sqrt cot x^2} \end{array}$

Toolbox:
• According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
• $\; \large \frac{d(cotx)}{dx} $$=-cosec^2x • \; \large \frac{d\sqrt x}{dx}$$ = \large \frac{1}{2 \sqrt x}$
Given $y = \;2\sqrt {cot (x^2)}$. This is of the form $y = f(g(x)$, where $g(x) = \cot x^2$
According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
$\textbf{Step 1}$:
While $g(x) = \cot x^3$, this itself is of the form $g(x) = f(h(x)$, where $h(x) = x^2$.
We can apply the Chain rule. According to the Chain Rule for differentiation, given two functions $f(x)$ and $h(x)$, and $y=f(h(x)) \rightarrow y' = f'(h(x)).h'(x)$.
$\Rightarrow h'(x) = 2\;x$
$\; \large \frac{d(cotx)}{dx} $$=-cosec^2x \Rightarrow f'(h(x)) = - cosec^2(x^2) \Rightarrow g'(x) = f'(h(x)).h'(x) = -2\;x\;cosec^2 x^2 \textbf{Step 2}: \; \large \frac{d\sqrt x}{dx}$$ = \large \frac{1}{2 \sqrt x}$
$\Rightarrow f'(g(x)) = 2 \; \large \frac{1}{2 \sqrt cot x^2} $$= \large \frac{1}{\sqrt cot x^2} \Rightarrow y' = f'(g(x)).g'(x) = \large \frac{1}{\sqrt cot x^2}$$ -2\;x\;cosec^2 x^2 = \large \frac{-2\;x\;cosec^2x^2}{\sqrt cot x^2}$