If f(x)=frac{(x+1) sinh x}{e^{2 x} tan x} and | vedclass

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(C) Given $f(x) = \frac{(x+1) \sinh x}{e^{2x} 	an x} = (x+1) \sinh x e^{-2x} \cot x$.Taking the natural logarithm on both sides: $\ln|f(x)| = \ln|x+1

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$-2(1+\operatorname{cosec} 2 x)$

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(C) Given $f(x) = \frac{(x+1) \sinh x}{e^{2x} 	an x} = (x+1) \sinh x e^{-2x} \cot x$.Taking the natural logarithm on both sides: $\ln|f(x)| = \ln|x+1| + \ln|\sinh x| - 2x + \ln|\cot x|$.Differentiating with respect to $x$: $\frac{f'(x)}{f(x)} = \frac{d}{dx}(\ln|x+1|) + \frac{d}{dx}(\ln|\sinh x|) + \frac{d}{dx}(-2x) + \frac{d}{dx}(\ln|\cot x|)$.$\frac{f'(x)}{f(x)} = \frac{1}{x+1} + \frac{\cosh x}{\sinh x} - 2 + \frac{-\operatorname{cosec}^2 x}{\cot x}$.Since $\frac{\cosh x}{\sinh x} = \operatorname{coth} x$ and $\frac{\operatorname{cosec}^2 x}{\cot x} = \frac{1}{\sin^2 x} \cdot \frac{\sin x}{\cos x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x} = 2 \operatorname{cosec} 2x$.Thus, $\frac{f'(x)}{f(x)} = \frac{1}{x+1} + \operatorname{coth} x - 2 - 2 \operatorname{cosec} 2x$.Comparing this with the given expression $\frac{1}{x+1} + \operatorname{coth} x + g(x)$, we get $g(x) = -2 - 2 \operatorname{cosec} 2x = -2(1 + \operatorname{cosec} 2x)$.

If $f(x)=\frac{(x+1) \sinh x}{e^{2 x} \tan x}$ and $\frac{f^{\prime}(x)}{f(x)}=\frac{1}{x+1}+\operatorname{coth} x+g(x)$, then $g(x)=$

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