## Condition of the problem

Find the differential $d y$.

$$
y=\arccos \left(\frac{1}{\sqrt{1+2 x^{2}}}\right), x>0
$$
## Solution

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

## Problem Kuznetsov Differentiation 4-5