Problem condition

Find the derivative.

$y=\frac{1}{2 \sqrt{2}}(\sin (\ln x)-(\sqrt{2}-1) \cdot \cos (\ln x)) x^{\sqrt{2}+1}$
## Solution

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

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

## Problem Kuznetsov Differentiation 15-3