Question

If $y = \sin^n x\, \cos\, nx$, then $\frac {dy}{dx}$ is

• A. $n\, sin^{n-1} x\,cos (n + 1)\, x$
• B. $n\, sin^{n-1} x\,sin (n + 1)\, x$
• C. $n\, sin^{n-1} x\,cos (n - 1)\, x$
• D. $n\, sin^{n-1} x \,cos\, nx$

Answer 1

Answer: (A)

Given, $y=\sin ^{n} x \cos n x$
$\frac{d y}{d x}=n \sin ^{n-1} x \cos x \cos n x-n \sin ^{n} x \sin n x$
$=n \sin ^{n-1} x[\cos x \cos n x-\sin x \sin n x]$
$=n \sin ^{n-1} x \cos (n+1) x$