## Problem Statement

Find the derivative.
$y=\frac{1}{2} \ln \frac{1+\cos x}{1-\cos x}-\frac{1}{\cos x}-\frac{1}{3 \cos ^{3} x}$
## Solution

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

Kuznetsov Differentiation 15-30