If y=log_{sin x} tan x,then left(frac{dy}{dx} | vedclass

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(C) Given $y = \log_{\sin x} 	an x$. Using the change of base formula,we can write $y = \frac{\log 	an x}{\log \sin x}$.Applying the quotient rule $

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$\frac{-4}{\log 2}$

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(C) Given $y = \log_{\sin x} 	an x$. Using the change of base formula,we can write $y = \frac{\log 	an x}{\log \sin x}$.Applying the quotient rule $\frac{d}{dx} \left( \frac{u}{v} ight) = \frac{v u' - u v'}{v^2}$,where $u = \log 	an x$ and $v = \log \sin x$:$u' = \frac{1}{	an x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = 2 \csc(2x)$.$v' = \frac{1}{\sin x} \cdot \cos x = \cot x$.Thus,$\frac{dy}{dx} = \frac{(\log \sin x)(2 \csc 2x) - (\log 	an x)(\cot x)}{(\log \sin x)^2}$.At $x = \frac{\pi}{4}$,$\sin x = \frac{1}{\sqrt{2}}$,$	an x = 1$,$\log 	an x = 0$,$\csc 2x = \csc \frac{\pi}{2} = 1$,and $\cot x = 1$.Substituting these values:$\left(\frac{dy}{dx}ight)_{x=\frac{\pi}{4}} = \frac{(\log \frac{1}{\sqrt{2}})(2 \cdot 1) - (0)(1)}{(\log \frac{1}{\sqrt{2}})^2} = \frac{2 \log(2^{-1/2})}{(\log 2^{-1/2})^2} = \frac{2 \cdot (-\frac{1}{2} \log 2)}{(-\frac{1}{2} \log 2)^2} = \frac{-\log 2}{\frac{1}{4} (\log 2)^2} = \frac{-4}{\log 2}$.

If $y=\log_{\sin x} \tan x$,then $\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}}$ has the value

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