differentiation

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



1) $n{\sin ^{n - 1}}x\cos (n + 1)x$
2) $n{\sin ^{n - 1}}x\cos \,nx$
3) $n{\sin ^{n - 1}}x\cos (n - 1)x$
4) $n{\sin ^{n - 1}}x\sin (n + 1)x$
Solution: Explanation: Explanation
standard-differentiation

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