If f(x) = frac{1+sec x}{2(sec x-1)} for 0

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(D) Given $f(x) = \frac{1+\sec x}{2(\sec x-1)}$. Simplifying $f(x)$ by converting to $\cos x$: $f(x) = \frac{1 + \frac{1}{\cos x}}{2(\frac{1}{\cos x}

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$-2 \operatorname{cosec} x$

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(D) Given $f(x) = \frac{1+\sec x}{2(\sec x-1)}$. Simplifying $f(x)$ by converting to $\cos x$: $f(x) = \frac{1 + \frac{1}{\cos x}}{2(\frac{1}{\cos x} - 1)} = \frac{\frac{\cos x + 1}{\cos x}}{2(\frac{1 - \cos x}{\cos x})} = \frac{1 + \cos x}{2(1 - \cos x)}$. Using the half-angle formulas $1 + \cos x = 2 \cos^2(\frac{x}{2})$ and $1 - \cos x = 2 \sin^2(\frac{x}{2})$: $f(x) = \frac{2 \cos^2(\frac{x}{2})}{2(2 \sin^2(\frac{x}{2}))} = \frac{1}{2} \cot^2(\frac{x}{2})$. Now, differentiate $f(x)$ with respect to $x$: $f^{\prime}(x) = \frac{1}{2} \cdot 2 \cot(\frac{x}{2}) \cdot (-\operatorname{cosec}^2(\frac{x}{2})) \cdot \frac{1}{2} = -\frac{1}{2} \cot(\frac{x}{2}) \operatorname{cosec}^2(\frac{x}{2})$. We know $f^{\prime}(x) = f(x) \cdot g(x)$, so $g(x) = \frac{f^{\prime}(x)}{f(x)}$. $g(x) = \frac{-\frac{1}{2} \cot(\frac{x}{2}) \operatorname{cosec}^2(\frac{x}{2})}{\frac{1}{2} \cot^2(\frac{x}{2})} = -\frac{\operatorname{cosec}^2(\frac{x}{2})}{\cot(\frac{x}{2})} = -\frac{1}{\sin^2(\frac{x}{2})} \cdot \frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})} = -\frac{1}{\sin(\frac{x}{2}) \cos(\frac{x}{2})}$. Multiplying numerator and denominator by $2$: $g(x) = -\frac{2}{2 \sin(\frac{x}{2}) \cos(\frac{x}{2})} = -\frac{2}{\sin x} = -2 \operatorname{cosec} x$.

If $f(x) = \frac{1+\sec x}{2(\sec x-1)}$ for $0 < x < \frac{\pi}{2}$ and $f^{\prime}(x) = f(x) \cdot g(x)$, then $g(x) =$

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