The value of log _{e} 2 cdot frac{d}{dx}(log | vedclass

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(D) Let $y = \log _{\cos x} \operatorname{cosec} x$. Using the change of base formula for logarithms,$y = \frac{\ln(\operatorname{cosec} x)}{\ln(\cos

Vedclass · Accepted Answer

$4$

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(D) Let $y = \log _{\cos x} \operatorname{cosec} x$. Using the change of base formula for logarithms,$y = \frac{\ln(\operatorname{cosec} x)}{\ln(\cos x)} = \frac{-\ln(\sin x)}{\ln(\cos x)}$. Differentiating with respect to $x$ using the quotient rule: $\frac{dy}{dx} = -\left[ \frac{\ln(\cos x) \cdot \frac{d}{dx}(\ln(\sin x)) - \ln(\sin x) \cdot \frac{d}{dx}(\ln(\cos x))}{(\ln(\cos x))^2} ight]$. $\frac{dy}{dx} = -\left[ \frac{\ln(\cos x) \cdot \cot x - \ln(\sin x) \cdot (-	an x)}{(\ln(\cos x))^2} ight]$. At $x = \frac{\pi}{4}$,$\sin x = \cos x = \frac{1}{\sqrt{2}}$,so $\ln(\sin x) = \ln(\cos x) = \ln(2^{-1/2}) = -\frac{1}{2} \ln 2$. Substituting these values: $\left. \frac{dy}{dx} ight|_{x=\frac{\pi}{4}} = -\left[ \frac{(-\frac{1}{2} \ln 2)(1) - (-\frac{1}{2} \ln 2)(-1)}{(-\frac{1}{2} \ln 2)^2} ight] = -\left[ \frac{-\frac{1}{2} \ln 2 - \frac{1}{2} \ln 2}{\frac{1}{4} (\ln 2)^2} ight] = -\left[ \frac{-\ln 2}{\frac{1}{4} (\ln 2)^2} ight] = \frac{4}{\ln 2}$. Finally,the value of $\log _{e} 2 \cdot \frac{dy}{dx}$ at $x = \frac{\pi}{4}$ is $\ln 2 \cdot \frac{4}{\ln 2} = 4$.

The value of $\log _{e} 2 \cdot \frac{d}{dx}(\log _{\cos x} \operatorname{cosec} x)$ at $x=\frac{\pi}{4}$ is.

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