## Task Condition

Find the derivative.

$$
y=-\frac{1}{3 \sin ^{3} x}-\frac{1}{\sin x}+\frac{1}{2} \ln \frac{1+\sin x}{1-\sin x}
$$
## Solution

$y^{\prime}=\left(-\frac{1}{3 \sin ^{3} x}-\frac{1}{\sin x}+\frac{1}{2} \ln \frac{1+\sin x}{1-\sin x}\right)^{\prime}=$

$=-\frac{-3}{3 \sin ^{4} x} \cdot \cos x-\frac{-1}{\sin ^{2} x} \cdot \cos x+$

$+\frac{1}{2} \cdot \frac{1-\sin x}{1+\sin x} \cdot \frac{\cos x \cdot(1-\sin x)-(1+\sin x) \cdot(-\cos x)}{(1-\sin x)^{2}}=$

$=\frac{\cos x}{\sin ^{4} x}+\frac{\cos x}{\sin ^{2} x}+\frac{\cos x-\cos x \cdot \sin x+\cos x+\sin x \cdot \cos x}{2\left(1-\sin ^{2} x\right)}=$

$=\frac{\cos ^{2} x}{\cos x \cdot \sin ^{4} x}+\frac{\cos ^{2} x \cdot \sin ^{2} x}{\cos x \cdot \sin ^{4} x}+\frac{1}{\cos x}=\frac{\cos ^{2} x+\cos ^{2} x \cdot \sin ^{2} x}{\cos x \cdot \sin ^{4} x}+\frac{\sin ^{4} x}{\cos x \cdot \sin ^{4} x}=$

$=\frac{\cos ^{2} x+\cos ^{2} x \cdot \sin ^{2} x+\sin ^{4} x}{\cos x \cdot \sin ^{4} x}=\frac{\cos ^{2} x+\sin ^{2} x \cdot\left(\cos ^{2} x+\sin ^{2} x\right)}{\cos x \cdot \sin ^{4} x}=$

$=\frac{\cos ^{2} x+\sin ^{2} x}{\cos x \cdot \sin ^{4} x}=\frac{1}{\cos x \cdot \sin ^{4} x}$