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If $y=\log _{\cos x} \sin x$, then $\frac{d y}{d x}$ is equal to
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Verified Answer
The correct answer is:
$\frac{(\cot x \log \cos x+\tan x \log \sin x)}{(\log \cos x)^{2}}$
Given, $y=\log _{\cos x} \sin x=\frac{\log \sin x}{\log \cos x}$
On differentiating w.r.t. $x$, we get
$$
\frac{d y}{d x}=\frac{\cot x \cdot \log \cos x+\tan x \cdot \log \sin x}{(\log \cos x)^{2}}
$$
On differentiating w.r.t. $x$, we get
$$
\frac{d y}{d x}=\frac{\cot x \cdot \log \cos x+\tan x \cdot \log \sin x}{(\log \cos x)^{2}}
$$
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