## Task Condition

Find the differential $d y$.

$$
y=\ln \left(\cos ^{2} x+\sqrt{1+\cos ^{4} x}\right)
$$
## Solution

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

## Problem Kuznetsov Differentiation 4-9