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

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

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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 $y = \frac{\log(	an x)}{\log(\sin x)}$.Differentiating with respect to $x$ using the quotient rule $\left( \frac{u}{v} ight)' = \frac{u'v - uv'}{v^2}$:$\frac{dy}{dx} = \frac{(\log(\sin x)) \cdot \frac{d}{dx}(\log(	an x)) - (\log(	an x)) \cdot \frac{d}{dx}(\log(\sin x))}{(\log(\sin x))^2}$.Since $\frac{d}{dx}(\log(	an x)) = \frac{\sec^2 x}{	an x} = \frac{1}{\sin x \cos x}$ and $\frac{d}{dx}(\log(\sin x)) = \cot x = \frac{\cos x}{\sin x}$,At $x = \frac{\pi}{4}$,$\sin x = \frac{1}{\sqrt{2}}$,$\cos x = \frac{1}{\sqrt{2}}$,$	an x = 1$,$\log(	an x) = 0$,and $\log(\sin x) = \log(2^{-1/2}) = -\frac{1}{2} \log 2$.Substituting these values:$\left( \frac{dy}{dx} ight)_{\pi/4} = \frac{(-\frac{1}{2} \log 2) \cdot (\frac{1}{(1/\sqrt{2})(1/\sqrt{2})}) - (0) \cdot (1)}{(-\frac{1}{2} \log 2)^2} = \frac{-\frac{1}{2} \log 2 \cdot 2}{\frac{1}{4} (\log 2)^2} = \frac{-\log 2}{\frac{1}{4} (\log 2)^2} = \frac{-4}{\log 2}$.

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

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