Question

$\frac{d\left[{\sin }^n\left(x\right)\cos \left(nx\right)\right]}{dx}=$

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

Answer 1

The correct option is A

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

Explanation for the correct option :

Given, $\frac{d\left[{\sin }^n\left(x\right)\cos \left(nx\right)\right]}{dx}$

For differentiating this apply theorem for product $\frac{d\left(u\times v\right)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$

$$\begin{gathered} \Rightarrow \frac{d\left[{\sin }^n\left(x\right)\cos \left(nx\right)\right]}{dx}=n{\sin }^{n-1}\left(x\right)\cos \left(x\right)\cos \left(nx\right)+{\sin }^n\left(x\right)\left(-\sin \left(nx\right)\right)n \\ \Rightarrow \frac{d\left[{\sin }^n\left(x\right)\cos \left(nx\right)\right]}{dx}=n{\sin }^{n-1}\left(x\right)\left[\cos \left(x\right)\cos \left(nx\right)-\sin \left(x\right)\left(\sin \left(nx\right)\right)\right] \\ \Rightarrow \frac{d\left[{\sin }^n\left(x\right)\cos \left(nx\right)\right]}{dx}=n{\sin }^{n-1}\left(x\right)\cos \left(x+nx\right) \\ \Rightarrow \frac{d\left[{\sin }^n\left(x\right)\cos \left(nx\right)\right]}{dx}=n{\sin }^{n-1}\left(x\right)\cos \left(\left(1+n\right)x\right) \end{gathered}$$

Hence, correct answer is option $\left(A\right)$