\[\frac{d}{dx}[{{\sin }^{n}}x\cos \,nx]=\]

1.	\[\frac{d}{dx}[{{\sin }^{n}}x\cos \,nx]=\]
A.	\[n{{\sin }^{n-1}}x\cos (n+1)x\]
B.	\[n{{\sin }^{n-1}}x\cos \,nx\]
C.	\[n{{\sin }^{n-1}}x\cos (n-1)x\]
D.	\[n{{\sin }^{n-1}}x\sin (n+1)x\]
Answer» B. \[n{{\sin }^{n-1}}x\cos \,nx\]

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\[\frac{d}{dx}[{{\sin }^{n}}x\cos \,nx]=\]

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