If y = log_{cos x} sin x,then frac{dy}{dx} is | vedclass

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(A) Given $y = \log_{\cos x} \sin x$. Using the change of base formula,$\log_a b = \frac{\log b}{\log a}$,we can write: $y = \frac{\log \sin x}{\log \

Vedclass · Accepted Answer

$\frac{\cot x \log \cos x + 	an x \log \sin x}{(\log \cos x)^2}$

Vedclass · Answer

(A) Given $y = \log_{\cos x} \sin x$. Using the change of base formula,$\log_a b = \frac{\log b}{\log a}$,we can write: $y = \frac{\log \sin x}{\log \cos x}$. Now,differentiate with respect to $x$ using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} ight) = \frac{v u' - u v'}{v^2}$: $\frac{dy}{dx} = \frac{(\log \cos x) \frac{d}{dx}(\log \sin x) - (\log \sin x) \frac{d}{dx}(\log \cos x)}{(\log \cos x)^2}$. Calculating the derivatives: $\frac{d}{dx}(\log \sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$. $\frac{d}{dx}(\log \cos x) = \frac{1}{\cos x} \cdot (-\sin x) = -	an x$. Substituting these back: $\frac{dy}{dx} = \frac{(\log \cos x)(\cot x) - (\log \sin x)(-	an x)}{(\log \cos x)^2}$. $\frac{dy}{dx} = \frac{\cot x \log \cos x + 	an x \log \sin x}{(\log \cos x)^2}$. Thus,the correct option is $A$.

If $y = \log_{\cos x} \sin x$,then $\frac{dy}{dx}$ is equal to

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