A) \[\frac{{{e}^{2x}}[(2x-1)\cot x-x\,\text{cose}{{\text{c}}^{2}}x]}{{{x}^{2}}}\]
B) \[\frac{{{e}^{2x}}[(2x+1)\cot x-x\,\text{cose}{{\text{c}}^{2}}x]}{{{x}^{2}}}\]
C) \[\frac{{{e}^{2x}}[(2x-1)\cot x+x\,\text{cose}{{\text{c}}^{2}}x]}{{{x}^{2}}}\]
D) None of these
Correct Answer: A
Solution :
\[y=\frac{{{e}^{2x}}\cos x}{x\sin x}\]Þ \[\log y=2x+\log \cos x-\log x-\log \sin x\] \[\frac{1}{y}\frac{dy}{dx}=2+\left( \frac{-\sin x}{\cos x} \right)-\frac{1}{x}-\frac{\cos x}{\sin x}\] Þ \[\frac{dy}{dx}={{e}^{2x}}\left[ \frac{2}{x}\cot x-\frac{1}{x}-\frac{1}{{{x}^{2}}}\cot x-\frac{{{\cot }^{2}}x}{x} \right]\] \[=\frac{{{e}^{2x}}}{{{x}^{2}}}[(2x-1)\cot x-x\,\text{cose}{{\text{c}}^{2}}x]\].
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