If y = e^ 2x x x x , then dy dx = - Vidyadip

MCQ

If $y = {{{e^{2x}}\cos x} \over {x\sin x}},$ then ${{dy} \over {dx}} = $

✓
${{{e^{2x}}[(2x - 1)\cot x - x\,{\rm{cose}}{{\rm{c}}^2}x]} \over {{x^2}}}$
B
${{{e^{2x}}[(2x + 1)\cot x - x\,{\rm{cose}}{{\rm{c}}^2}x]} \over {{x^2}}}$
C
${{{e^{2x}}[(2x - 1)\cot x + x\,{\rm{cose}}{{\rm{c}}^2}x]} \over {{x^2}}}$
D
None of these

✓

Answer

Correct option: A.

${{{e^{2x}}[(2x - 1)\cot x - x\,{\rm{cose}}{{\rm{c}}^2}x]} \over {{x^2}}}$

a
(a) $y = \frac{{{e^{2x}}\cos x}}{{x\sin x}}$

==> $\log y = 2x + \log \cos x - \log x - \log \sin x$

$\frac{1}{y}\frac{{dy}}{{dx}} = 2 + \left( {\frac{{ - \sin x}}{{\cos x}}} \right) - \frac{1}{x} - \frac{{\cos x}}{{\sin x}}$

==> $\frac{{dy}}{{dx}} = {e^{2x}}\left[ {\frac{2}{x}\cot x - \frac{1}{x} - \frac{1}{{{x^2}}}\cot x - \frac{{{{\cot }^2}x}}{x}} \right]$

$ = \frac{{{e^{2x}}}}{{{x^2}}}[(2x - 1)\cot x - x\,{\rm{cose}}{{\rm{c}}^2}x]$.

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