A) \[\frac{\cos x}{2y-1}\]
B) \[\frac{\sin x}{2y-1}\]
C) \[\frac{\log x}{1-2y}\]
D) \[\frac{\sin x}{1+xy}\]
Correct Answer: A
Solution :
\[y=\sqrt{\sin x+y}\] \[\Rightarrow \] \[{{y}^{2}}=sin\text{ }x+y\] Differentiating w.r.t.\['x',\] \[2y\frac{dy}{dx}=\cos x+\frac{dy}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}(2y-1)=\cos x\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{\cos x}{2y-1}\]
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