$\frac{x^{5}-\cos x}{\sin x}$

solution:

Given $y=\frac{x^{5}-\cos x}{\sin x}$

$\mathrm{d} / \mathrm{dx}\left(\mathrm{x}^{5}-\cos \mathrm{x}\right) / \sin \mathrm{x}=\left[\sin \mathrm{x} \cdot \mathrm{d} / \mathrm{dx}\left(\mathrm{x}^{5}-\cos \right.\right.$ $\left.x)-\left(x^{5}-\cos x\right) \cdot d / d x(\sin x)\right] / \sin ^{2} x$

By using quotient rule,

$=\left[\sin x\left(5 x^{4}+\sin x\right)-\right.$ $\left.\left(x^{5}-\cos x\right)(\cos x)\right] / \sin ^{2} x$

$=\left[5 x^{4} \cdot \sin x+\sin ^{2} x-\right.$ $\left.x^{5} \cos x+\cos ^{2} x\right] / \sin ^{2} x$

$=\left[5 x^{4} \sin x-x^{5} \cos x\right.$ $\left.+\left(\sin ^{2}+\cos ^{2} x\right)\right] / \sin ^{2} x$

$=\left[5 x^{4} \sin x-x^{5} \cos x\right.$ $+1] / \sin ^{2} x$

Hence, the required answer is $\left[5 x^{4} \sin x-x^{5}\right.$ $\cos x+1] / \sin ^{2} x$