The derivative of $\log {\log [\log (\sin x)]}$ w.r.t. sin x is

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$\frac{- 1}{\sin x \cdot \log \sin x \cdot \log \log \sin x}$

$\frac{1}{\log \sin x \log \log \sin x}$

$\frac{- 1}{\log \sin x \log \log \sin x}$

$\frac{1}{\sin x \log \sin x \log \log \sin x}$

answer is A.

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Detailed Solution

$N o w \frac{d}{d x} (\log \{\log (\log (\sin x))\} = \frac{1}{\log \log \sin x} \cdot \frac{d}{d x} (\log (\log (\sin x))) = \frac{1}{\log \log \sin x} \frac{1}{\log \sin x} \frac{d}{d x} (\log (\sin x)) = \frac{1}{\log \log \sin x} \cdot \frac{1}{\log \sin x} \cdot \frac{1}{\sin x} \cdot \cos x Now required derivative = \frac{\frac{1}{\log \log \sin x} \cdot \frac{1}{\log \sin x} \cdot \frac{1}{\sin x} \cdot \cos x}{\cos x} = \frac{1}{\sin x \cdot \log \sin x \cdot \log \log \sin x}$