## Problem Statement

Find the derivative.

$$
y=x \ln (\sqrt{1-x}+\sqrt{1+x})+\frac{1}{2}(\arcsin x-x)
$$
## Solution

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

$$
\begin{aligned}
& =\ln (\sqrt{1-x}+\sqrt{1+x})+x \cdot \frac{1-x-2 \sqrt{1-x^{2}}+1+x}{-4 x \sqrt{1-x^{2}}}+\frac{1-\sqrt{1-x^{2}}}{2 \sqrt{1-x^{2}}}= \\
& =\ln (\sqrt{1-x}+\sqrt{1+x})-\frac{1-\sqrt{1-x^{2}}}{2 \sqrt{1-x^{2}}}+\frac{1-\sqrt{1-x^{2}}}{2 \sqrt{1-x^{2}}}= \\
& =\ln (\sqrt{1-x}+\sqrt{1+x})
\end{aligned}
$$

## Problem Kuznetsov Differentiation 14-20