A) \[\frac{y}{1-y}\]
B) \[\frac{1}{1-y}\]
C) \[\frac{y}{1+y}\]
D) \[\frac{y}{y-1}\]
Correct Answer: A
Solution :
\[y={{e}^{x+y}}\] Þ \[\log y=(x+y)\log e\] Þ \[\frac{1}{y}\frac{dy}{dx}=1+\frac{dy}{dx}\] Þ \[={{\sin }^{2}}\alpha +\frac{1}{2}(\cos 2\alpha +\cos 2\beta )\].
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