## Task Condition

Find the derivative.

$$
y=\frac{2}{3}\left(4 x^{2}-4 x+3\right) \sqrt{x^{2}-x}+(2 x-1)^{4} \cdot \arcsin \frac{1}{2 x-1}, 2 x-1>0
$$
## Solution

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

Problem Kuznetsov Differentiation 13-21