d dx [ ^n x ,nx] = - Vidyadip

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
$n{\sin ^{n - 1}}x\sin (n + 1)x$

✓

Answer

Correct option: A.

$n{\sin ^{n - 1}}x\cos (n + 1)x$

a
(a) $\frac{d}{{dx}}[{\sin ^n}x\cos nx] = n{\sin ^{n - 1}}x\cos x\cos nx - n\sin nx{\sin ^n}x$

$ = 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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