## Problem Statement

Find the derivative.

$$
y=2 \frac{\cos x}{\sin ^{4} x}+3 \frac{\cos x}{\sin ^{2} x}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=2 \frac{(\cos x)^{\prime} \cdot \sin ^{4} x-\cos x\left(\sin ^{4} x\right)^{\prime}}{\sin ^{8} x}+3 \frac{(\cos x)^{\prime} \cdot \sin ^{2} x-\cos x\left(\sin ^{2} x\right)^{\prime}}{\sin ^{4} x}= \\
& =2 \frac{-\sin ^{5} x-4 \sin ^{3} x \cdot \cos ^{2} x}{\sin ^{8} x}+3 \frac{-\sin ^{3} x-2 \sin x \cdot \cos ^{2} x}{\sin ^{4} x}=2 \frac{-\sin ^{2} x-4 \cos ^{2} x}{\sin ^{5} x}+3 \frac{-\sin ^{2} x-2 \cos ^{2} x}{\sin ^{3} x}= \\
& =\frac{-2 \sin ^{2} x-8\left(1-\sin ^{2} x\right)}{\sin ^{5} x}+\frac{-3 \sin ^{2} x-6\left(1-\sin ^{2} x\right)}{\sin ^{3} x}= \\
& =\frac{-2 \sin ^{2} x-8+8 \sin ^{2} x}{\sin ^{5} x}+\frac{-3 \sin ^{2} x-6+6 \sin ^{2} x}{\sin ^{3} x}=\frac{6 \sin ^{2} x-8}{\sin ^{5} x}+\frac{3 \sin ^{2} x-6}{\sin ^{3} x}=\frac{6 \sin ^{2} x-8+3 \sin ^{4} x-6 \sin ^{2} x}{\sin ^{5} x}= \\
& =\frac{3}{\sin x}-\frac{8}{\sin ^{5} x}=3 \operatorname{cosec} x-8 \operatorname{cosec}^{5} x
\end{aligned}
$$

## Problem Kuznetsov Differentiation 15-27