## Task Condition

Find the derivative.

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

$$
\begin{aligned}
& y^{\prime}=\left(\frac{2 x^{2}-x-1}{3 \sqrt{2+4 x}}\right)^{\prime}=\frac{(4 x-1) \cdot \sqrt{2+4 x}-\left(2 x^{2}-x-1\right) \cdot \frac{1}{2 \sqrt{2+4 x}} \cdot 4}{3(2+4 x)}= \\
& =\frac{(4 x-1) \cdot(2+4 x)-\left(2 x^{2}-x-1\right) \cdot \frac{1}{3}}{3(2+4 x) \sqrt{2+4 x}}=\frac{8 x+10 x^{2}-2-4 x-4 x^{2}+2 x+2}{3(2+4 x) \sqrt{2+4 x}}= \\
& =\frac{12 x^{2}+6 x}{6(1+2 x) \sqrt{2+4 x}}=\frac{6 x(2 x+1)}{6(1+2 x) \sqrt{2+4 x}}=\frac{x}{\sqrt{2+4 x}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 6-4