If A = 2^x x x , then dA dx = - Vidyadip

MCQ

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

✓
${{{2^{x - 1}}\left\{ { - 2x\,{\rm{cos}}{\rm{e}}{{\rm{c}}^2}x + \cot x.\log \left( {{{{4^x}} \over e}} \right)} \right\}} \over {{x^{3/2}}}}$
B
${{{2^{x - 1}}\left\{ { - 2x\cos {\rm{e}}{{\rm{c}}^2}x + \cot x.\log \left( {{{{4^x}} \over e}} \right)} \right\}} \over x}$
C
${{2x\left\{ { - 2x{\rm{cose}}{{\rm{c}}^2}x + \cot x.\log \left( {{{{4^x}} \over e}} \right)} \right\}} \over {{x^{{\rm{3/2}}}}}}$
D
None of these

✓

Answer

Correct option: A.

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

a
(a) $\frac{{dA}}{{dx}} = \frac{{\sqrt x \{ {2^x}{{\log }_e}2\cot x - {2^x}{\rm{cose}}{{\rm{c}}^2}x\} - {2^x}\cot x\frac{1}{{2\sqrt x }}}}{x}$

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

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