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

Question

Accepted Answer

$\frac{{{2^{x - 1}}\left\{ { - 2x\,{\rm{cos}}{\rm{e}}{{\rm{c}}^2}x + \cot x.\log \left( {\frac{{{4^x}}}{e}} \right)} \right\}}}{{{x^{3/2}}}}$

Answer

$\frac{{{2^{x - 1}}\left\{ { - 2x\cos {\rm{e}}{{\rm{c}}^2}x + \cot x.\log \left( {\frac{{{4^x}}}{e}} \right)} \right\}}}{x}$

Answer

$\frac{{2x\left\{ { - 2x{\rm{cose}}{{\rm{c}}^2}x + \cot x.\log \left( {\frac{{{4^x}}}{e}} \right)} \right\}}}{{{x^{{\rm{3/2}}}}}}$

Answer

None of these

differentiation