If f(x) = frac{e^{-x} sin x}{log_e x} and f'( | vedclass

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(C) Given $f(x) = \frac{e^{-x} \sin x}{\log_e x}$.Taking the natural logarithm on both sides: $\ln f(x) = \ln(e^{-x}) + \ln(\sin x) - \ln(\ln x) = -x

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$2e^{-2} - \operatorname{cosec}^2(e)$

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(C) Given $f(x) = \frac{e^{-x} \sin x}{\log_e x}$.Taking the natural logarithm on both sides: $\ln f(x) = \ln(e^{-x}) + \ln(\sin x) - \ln(\ln x) = -x + \ln(\sin x) - \ln(\ln x)$.Differentiating with respect to $x$: $\frac{f'(x)}{f(x)} = -1 + \frac{\cos x}{\sin x} - \frac{1}{x \ln x}$.Since $f'(x) = f(x) \cdot g(x)$, we have $g(x) = \frac{f'(x)}{f(x)} = -1 + \cot x - \frac{1}{x \ln x}$.Differentiating $g(x)$ with respect to $x$: $g'(x) = -\operatorname{cosec}^2 x - \frac{d}{dx} \left( \frac{1}{x \ln x} \right) = -\operatorname{cosec}^2 x - \left( \frac{-(1 \cdot \ln x + x \cdot \frac{1}{x})}{(x \ln x)^2} \right) = -\operatorname{cosec}^2 x + \frac{\ln x + 1}{x^2 (\ln x)^2}$.At $x = e$: $g'(e) = -\operatorname{cosec}^2(e) + \frac{\ln e + 1}{e^2 (\ln e)^2} = -\operatorname{cosec}^2(e) + \frac{1 + 1}{e^2 (1)^2} = 2e^{-2} - \operatorname{cosec}^2(e)$.

If $f(x) = \frac{e^{-x} \sin x}{\log_e x}$ and $f'(x) = f(x) \cdot g(x)$, then $g'(e) =$

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