Question
यदि $y = {\log _{\cos x}}\sin x$, तो $\frac{{dy}}{{dx}} =$
$\therefore \frac{{dy}}{{dx}} = \frac{{\cot x.\log \cos x + (\log \sin x)\tan x}}{{{{(\log \cos x)}^2}}}$.
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$\tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}, x \neq 0$