A) \[\frac{\sin x}{2y-1}\]
B) \[\frac{\cos x}{2y-1}\]
C) \[\frac{\sin x}{2y+1}\]
D) \[\frac{\cos x}{2y+1}\]
Correct Answer: B
Solution :
\[y=\sqrt{\sin x+y},\] Þ \[{{y}^{2}}=\sin x+y\] Differentiate with respect to x, \[2y.\frac{dy}{dx}=\cos x+\frac{dy}{dx}\] Þ \[\frac{dy}{dx}(2y-1)=\cos x\]Þ \[\frac{dy}{dx}=\frac{\cos x}{2y-1}\].
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