## Problem Statement

Find the derivative.

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

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

$=\frac{(\arcsin x)^{\prime} \cdot \sqrt{1-x^{2}}-\arcsin x \cdot\left(\sqrt{1-x^{2}}\right)^{\prime}}{1-x^{2}}+\frac{1}{2} \cdot \frac{1+x}{1-x} \cdot \frac{-1 \cdot(1+x)-(1-x) \cdot 1}{(1+x)^{2}}=$

$=\frac{\frac{1}{\sqrt{1-x^{2}}} \cdot \sqrt{1-x^{2}}-\arcsin x \cdot \frac{1}{2 \sqrt{1-x^{2}}} \cdot 2 x}{1-x^{2}}+\frac{1}{2} \cdot \frac{1}{1-x} \cdot \frac{-2}{1+x}=$

$=\frac{1-\frac{x \cdot \arcsin x}{\sqrt{1-x^{2}}}}{1-x^{2}}-\frac{1}{1-x^{2}}=-\frac{x \cdot \arcsin x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}$

## Problem Kuznetsov Differentiation 14-31