MCQ
${d \over {dx}}[{\sin ^n}x\cos \,nx] = $
- ✓$n{\sin ^{n - 1}}x\cos (n + 1)x$
- B$n{\sin ^{n - 1}}x\cos \,nx$
- C$n{\sin ^{n - 1}}x\cos (n - 1)x$
- D$n{\sin ^{n - 1}}x\sin (n + 1)x$
$ = 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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