Question

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

• A. $nsi{n}^{n-1}xcos(n+1)x$
• B. $nsi{n}^{n-1}xsin(n+1)x$
• C. $nsi{n}^{n-1}xcos(n-1)x$
• D. $nsi{n}^{n-1}xcosnx$

Answer 1

Answer: (A)

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