A) \[\frac{-\sin \,4x}{4}\]
B) \[\frac{-\sin \,2x}{2}\]
C) \[\frac{\sin \,4x}{4}\]
D) \[\frac{-\sin \,4x}{2}\]
Correct Answer: D
Solution :
We have \[y={{\sin }^{2}}x+{{\cos }^{4}}x\] \[\therefore \] \[\frac{dy}{dx}=2\sin x\operatorname{cosx}+4co{{s}^{3}}x(-\sin \,x)\] \[=\sin 2x-4\sin x\cos x({{\cos }^{2}}x)\] \[=\sin 2x-2\sin 2x\left( \frac{\cos 2x+1}{2} \right)\] \[=\sin 2x-sin\,2x\,\cos \,2x-\,\sin \,2x\] \[=-\sin 2x\,\cos \,2x=\frac{-\sin \,4x}{2}\]
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