If \[A=\frac{{{2}^{x}}\cot x}{\sqrt{x}},\]then \[\frac{dA}{dx}=\]

1.	If \[A=\frac{{{2}^{x}}\cot x}{\sqrt{x}},\]then \[\frac{dA}{dx}=\]
A.	\[\frac{{{2}^{x-1}}\left\{ -2x\,\text{cos}\text{e}{{\text{c}}^{2}}x+\cot x.\log \left( \frac{{{4}^{x}}}{e} \right) \right\}}{{{x}^{3/2}}}\]
B.	\[\frac{{{2}^{x-1}}\left\{ -2x\cos \text{e}{{\text{c}}^{2}}x+\cot x.\log \left( \frac{{{4}^{x}}}{e} \right) \right\}}{x}\]
C.	\[\frac{2x\left\{ -2x\text{cose}{{\text{c}}^{2}}x+\cot x.\log \left( \frac{{{4}^{x}}}{e} \right) \right\}}{{{x}^{\text{3/2}}}}\]
D.	None of these
Answer» B. \[\frac{{{2}^{x-1}}\left\{ -2x\cos \text{e}{{\text{c}}^{2}}x+\cot x.\log \left( \frac{{{4}^{x}}}{e} \right) \right\}}{x}\]

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If \[A=\frac{{{2}^{x}}\cot x}{\sqrt{x}},\]then \[\frac{dA}{dx}=\]

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