Find the derivative of $\frac{x^{5}-\cos x}{\sin x}$.

Let $h(x) = \frac{x^{5}-\cos x}{\sin x}$. We use the quotient rule,$\left(\frac{u}{v}\right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^{2}}$,where $u = x^{5}-\cos x$ and $v = \sin x$.
$h^{\prime}(x) = \frac{\frac{d}{dx}(x^{5}-\cos x) \cdot \sin x - (x^{5}-\cos x) \cdot \frac{d}{dx}(\sin x)}{(\sin x)^{2}}$
$h^{\prime}(x) = \frac{(5x^{4} + \sin x) \sin x - (x^{5}-\cos x) \cos x}{\sin^{2} x}$
$h^{\prime}(x) = \frac{5x^{4} \sin x + \sin^{2} x - x^{5} \cos x + \cos^{2} x}{\sin^{2} x}$
Since $\sin^{2} x + \cos^{2} x = 1$,we have:
$h^{\prime}(x) = \frac{5x^{4} \sin x - x^{5} \cos x + 1}{\sin^{2} x}$