A) \[\frac{1}{2}\]
B) \[\frac{-1}{2}\]
C) -1
D) 0
Correct Answer: B
Solution :
[b] \[\because y=\frac{\cos n}{1+sinx}\] \[\therefore \frac{dy}{dx}=\frac{(1+\sin x)(-\sin x)-\cos x.(0+\cos x)}{{{(1+\sin x)}^{2}}}\] \[=\frac{\left( -\sin x-{{\sin }^{2}}x \right)-{{\cos }^{2}}x}{{{(1+\sin x)}^{2}}}=\frac{-(1+\sin x)}{{{(1+\sin x)}^{2}}}=\frac{-1}{(1+\sin x)}\] \[\therefore \,{{\left( \frac{dy}{dx} \right)}_{at\,x=\frac{1}{2}}}=\frac{-1}{1+\sin \,\frac{\pi }{2}\,}=\frac{-1}{2}\]
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