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 $y = \frac{\ln(	an x)}{\ln(\sin x)}$.Applying the quotient rule

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

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

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

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