## Problem Statement

Find the derivative.

$$
y=\ln \frac{1+\sqrt{-3+4 x-x^{2}}}{2-x}+\frac{2}{2-x} \sqrt{-3+4 x-x^{2}}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\ln \frac{1+\sqrt{-3+4 x-x^{2}}}{2-x}+\frac{2}{2-x} \sqrt{-3+4 x-x^{2}}\right)^{\prime}= \\
& =\frac{2-x}{1+\sqrt{-3+4 x-x^{2}}} \cdot \frac{\frac{1}{2 \sqrt{-3+4 x-x^{2}}} \cdot(4-2 x) \cdot(2-x)-\left(1+\sqrt{-3+4 x-x^{2}}\right) \cdot(-1)}{(2-x)^{2}}+ \\
& +\frac{2 \cdot \frac{1}{2 \sqrt{-3+4 x-x^{2}}} \cdot(4-2 x) \cdot(2-x)-2 \sqrt{-3+4 x-x^{2}} \cdot(-1)}{(2-x)^{2}}= \\
& =\frac{1}{1+\sqrt{-3+4 x-x^{2}}} \cdot \frac{(2-x)^{2}+\sqrt{-3+4 x-x^{2}}+\left(-3+4 x-x^{2}\right)}{(2-x) \sqrt{-3+4 x-x^{2}}}+ \\
& +\frac{2 \cdot(2-x)^{2}+2 \cdot\left(-3+4 x-x^{2}\right)}{(2-x)^{2} \cdot \sqrt{-3+4 x-x^{2}}}= \\
& =\frac{1}{1+\sqrt{-3+4 x-x^{2}}} \cdot \frac{4-4 x+x^{2}+\sqrt{-3+4 x-x^{2}}-3+4 x-x^{2}}{(2-x) \sqrt{-3+4 x-x^{2}}}+ \\
& +\frac{8-8 x+2 x^{2}-6+8 x-2 x^{2}}{(2-x)^{2} \cdot \sqrt{-3+4 x-x^{2}}}= \\
& =\frac{1}{1+\sqrt{-3+4 x-x^{2}}} \cdot \frac{1+\sqrt{-3+4 x-x^{2}}}{(2-x) \sqrt{-3+4 x-x^{2}}+\frac{2-x}{(2-x)^{2} \cdot \sqrt{-3+4 x-x^{2}}}=} \\
& =\frac{1}{(2-x) \sqrt{-3+4 x-x^{2}}+\frac{2-x}{(2-x)^{2} \cdot \sqrt{-3+4 x-x^{2}}}=} \\
& =\frac{2-x+2}{(2-x)^{2} \cdot \sqrt{-3+4 x-x^{2}}}=\frac{4-x}{(2-x)^{2} \cdot \sqrt{-3+4 x-x^{2}}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 13-15