$\frac{\mathrm{d}}{\mathrm{dx}}\left[{\sin }^{\mathrm{n}} \mathrm{x}·\sin \mathrm{nx}\right]=$

The given function is $y={\sin }^{n}x·\sin nx$

Use the product rule to find the derivative of the above function

$\frac{d}{dx}\left(U·V\right)=U·\frac{d}{dx}\left(V\right)+V·\frac{d}{dx}\left(U\right)$

Hence,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left({\sin }^{n}x·\sin nx\right)\\ & =& n{\sin }^{n-1}x\cos x\left(\sin nx\right)+{\sin }^{n}x\left(n\cos nx\right)\\ & =& n{\sin }^{n-1}x\left(\cos x·\sin nx+\sin x\cos nx\right)\\ & =& n {\sin }^{n-1}x\left(\sin \left(n+1\right)x\right)\end{array}$

Therefore, $\frac{\mathrm{d}}{\mathrm{dx}}\left[{\sin }^{\mathrm{n}} \mathrm{x}·\sin \mathrm{nx}\right]=\boxed{\mathbf{n}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{·}\left[sin\left(n+1\right)x\right]}$