A) \[\frac{{{y}^{2}}\cot x}{1-y\log \sin x}\]
B) \[\frac{{{y}^{2}}\cot x}{1+y\log \sin x}\]
C) \[\frac{y\cot x}{1-y\log \sin x}\]
D) \[\frac{y\cot x}{1+y\log \sin x}\]
Correct Answer: A
Solution :
\[y={{(\sin x)}^{{{(\sin x)}^{(\sin x).....\infty }}}}\] Þ \[y={{(\sin x)}^{y}}\Rightarrow {{\log }_{e}}y=y\log \sin x\] Þ \[\frac{1}{y}\frac{dy}{dx}=\frac{dy}{dx}[\log \sin x+y\cot x]\] \[\therefore \frac{dy}{dx}=\frac{{{y}^{2}}\cot x}{1-y\log \sin x}\].
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