Condition of the problem

Find the derivative.

$$
y=\operatorname{arctg}\left(\frac{\cos x}{\sqrt[4]{\cos 2 x}}\right)
$$
## Solution

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

## Problem Kuznetsov Differentiation $15-4$