Condition of the problem

Find the derivative.

$y=\ln ^{2}(x+\cos x)$
## Solution

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

## Problem Kuznetsov Differentiation 8-7