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) \[n{{\sin }^{n-1}}x\sin (n+1)x\]
Correct Answer: A
Solution :
\[\frac{d}{dx}[{{\sin }^{n}}x\cos nx]=n{{\sin }^{n-1}}x\cos x\cos nx-n\sin nx{{\sin }^{n}}x\] \[=n{{\sin }^{n-1}}x[\cos x\cos nx-\sin nx\sin x]=n{{\sin }^{n-1}}x\cos \,(n+1)x\].
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