Find the derivative of the following function: $\frac{x}{\sin^{n} x}$ (where $n$ is an integer).

Let $f(x) = \frac{x}{\sin^{n} x}$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
Here,$u = x$ and $v = \sin^{n} x$.
$\frac{du}{dx} = 1$ and $\frac{dv}{dx} = n \sin^{n-1} x \cos x$ (by chain rule).
Substituting these into the quotient rule formula:
$f'(x) = \frac{\sin^{n} x (1) - x (n \sin^{n-1} x \cos x)}{(\sin^{n} x)^2}$.
$f'(x) = \frac{\sin^{n} x - n x \sin^{n-1} x \cos x}{\sin^{2n} x}$.
Factoring out $\sin^{n-1} x$ from the numerator:
$f'(x) = \frac{\sin^{n-1} x (\sin x - n x \cos x)}{\sin^{2n} x}$.
Simplifying the expression:
$f'(x) = \frac{\sin x - n x \cos x}{\sin^{n+1} x}$.