## Task Condition

Find the derivative.

$y=\frac{1+x^{2}}{2 \sqrt{1+2 x^{2}}}$
## Solution

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

## Problem Kuznetsov Differentiation 6-13