## Problem Statement

Find the derivative.

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

$$
\begin{aligned}
& y^{\prime}=\left(\frac{2}{x-1} \cdot \sqrt{2 x-x^{2}}+\ln \frac{1+\sqrt{2 x-x^{2}}}{x-1}\right)^{\prime}= \\
& =\frac{\frac{2}{2 \sqrt{2 x-x^{2}}} \cdot(2-2 x)-2 \sqrt{2 x-x^{2}} \cdot 1}{(x-1)^{2}}+\frac{x-1}{1+\sqrt{2 x-x^{2}}} \cdot\left(\frac{1+\sqrt{2 x-x^{2}}}{x-1}\right)^{\prime}= \\
& =\frac{(2-2 x)-2\left(2 x-x^{2}\right)}{(x-1)^{2} \sqrt{2 x-x^{2}}}+\frac{x-1}{1+\sqrt{2 x-x^{2}}} \cdot \frac{\frac{1}{2 \sqrt{2 x-x^{2}}} \cdot(2-2 x) \cdot(x-1)-\left(1+\sqrt{2 x-x^{2}}\right) \cdot 1}{(x-1)^{2}}= \\
& =\frac{2-2 x-4 x+2 x^{2}}{(x-1)^{2} \sqrt{2 x-x^{2}}}+\frac{\frac{1}{\sqrt{2 x-x^{2}}} \cdot(1-x) \cdot(x-1)-1-\sqrt{2 x-x^{2}}}{\left(1+\sqrt{2 x-x^{2}}\right) \cdot(x-1)}= \\
& =\frac{2-6 x+2 x^{2}}{(x-1)^{2} \sqrt{2 x-x^{2}}}+\frac{-(x-1)^{2}-\sqrt{2 x-x^{2}}-\left(2 x-x^{2}\right)}{\left(1+\sqrt{2 x-x^{2}}\right) \cdot(x-1) \cdot \sqrt{2 x-x^{2}}}= \\
& =\frac{2-6 x+2 x^{2}}{(x-1)^{2} \sqrt{2 x-x^{2}}}-\frac{x^{2}-2 x+1+\sqrt{2 x-x^{2}}+2 x-x^{2}}{\left(1+\sqrt{2 x-x^{2}}\right) \cdot(x-1) \cdot \sqrt{2 x-x^{2}}}= \\
& =\frac{2-6 x+2 x^{2}}{(x-1)^{2} \sqrt{2 x-x^{2}}}-\frac{1+\sqrt{2 x-x^{2}}}{\left(1+\sqrt{2 x-x^{2}}\right) \cdot(x-1) \cdot \sqrt{2 x-x^{2}}}= \\
& =\frac{2-6 x+2 x^{2}}{(x-1)^{2} \sqrt{2 x-x^{2}}}-\frac{1}{(x-1) \cdot \sqrt{2 x-x^{2}}}=\frac{2-6 x+2 x^{2}}{(x-1)^{2} \sqrt{2 x-x^{2}}}-\frac{x-1}{(x-1)^{2} \cdot \sqrt{2 x-x^{2}}}= \\
& =\frac{2-6 x+2 x^{2}-x+1}{(x-1)^{2} \sqrt{2 x-x^{2}}}=\frac{2 x^{2}-7 x+3}{(x-1)^{2} \sqrt{2 x-x^{2}}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 13-5