If y = frac{e^{2x} cos x}{x sin x},then frac{ | vedclass

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(A) Given $y = \frac{e^{2x} \cos x}{x \sin x} = \frac{e^{2x} \cot x}{x}$.Taking the natural logarithm on both sides:$\ln y = \ln(e^{2x}) + \ln(\cot x)

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$\frac{e^{2x}[(2x - 1)\cot x - x \csc^2 x]}{x^2}$

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(A) Given $y = \frac{e^{2x} \cos x}{x \sin x} = \frac{e^{2x} \cot x}{x}$.Taking the natural logarithm on both sides:$\ln y = \ln(e^{2x}) + \ln(\cot x) - \ln(x) = 2x + \ln(\cot x) - \ln x$.Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = 2 + \frac{1}{\cot x} \cdot (-\csc^2 x) - \frac{1}{x}$.$\frac{1}{y} \frac{dy}{dx} = 2 - 	an x \cdot \csc^2 x - \frac{1}{x} = 2 - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} - \frac{1}{x} = 2 - \frac{1}{\sin x \cos x} - \frac{1}{x} = 2 - 2 \csc(2x) - \frac{1}{x}$.Alternatively,using the quotient rule on $y = \frac{e^{2x} \cot x}{x}$:$\frac{dy}{dx} = \frac{x \cdot \frac{d}{dx}(e^{2x} \cot x) - e^{2x} \cot x \cdot \frac{d}{dx}(x)}{x^2}$.$\frac{dy}{dx} = \frac{x [2e^{2x} \cot x + e^{2x} (-\csc^2 x)] - e^{2x} \cot x}{x^2}$.$\frac{dy}{dx} = \frac{e^{2x} [2x \cot x - x \csc^2 x - \cot x]}{x^2}$.$\frac{dy}{dx} = \frac{e^{2x} [(2x - 1) \cot x - x \csc^2 x]}{x^2}$.

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

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