If \[y=\frac{{{e}^{2x}}\cos x}{x\sin x},\]then \[\frac{dy}{dx}=\] [AI CBSE 1982]

\[\frac{{{e}^{2x}}[(2x+1)\cot x-x\,\text{cose}{{\text{c}}^{2}}x]}{{{x}^{2}}}\]

\[\frac{{{e}^{2x}}[(2x-1)\cot x-x\,\text{cose}{{\text{c}}^{2}}x]}{{{x}^{2}}}\]

\[\frac{{{e}^{2x}}[(2x-1)\cot x+x\,\text{cose}{{\text{c}}^{2}}x]}{{{x}^{2}}}\]

If \[y=\frac{{{e}^{2x}}\cos x}{x\sin x},\]then \[\frac{dy}{dx}=\] �

1.	If \[y=\frac{{{e}^{2x}}\cos x}{x\sin x},\]then \[\frac{dy}{dx}=\] [AI CBSE 1982]
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
Answer» B. \[\frac{{{e}^{2x}}[(2x+1)\cot x-x\,\text{cose}{{\text{c}}^{2}}x]}{{{x}^{2}}}\]

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If \[y=\frac{{{e}^{2x}}\cos x}{x\sin x},\]then \[\frac{dy}{dx}=\] [AI CBSE 1982]

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