## Task Condition

Find the derivative.

$$
y=\frac{(x+3) \sqrt{2 x-1}}{2 x+7}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{(x+3) \sqrt{2 x-1}}{2 x+7}\right)^{\prime}=\frac{\left(\sqrt{2 x-1}+(x+3) \frac{1}{2 \sqrt{2 x-1}} \cdot 2\right) \cdot(2 x+7)-(x+3) \sqrt{2 x-1} \cdot 2}{(2 x+7)^{2}}= \\
& =\frac{(2 x-1+(x+3)) \cdot(2 x+7)-(x+3) \cdot(2 x-1) \cdot 2}{(2 x+7)^{2} \sqrt{2 x-1}}= \\
& =\frac{(3 x+2) \cdot(2 x+7)-2 \cdot\left(2 x^{2}-x+6 x-3\right)}{(2 x+7)^{2} \sqrt{2 x-1}}= \\
& =\frac{6 x^{2}+21 x+4 x+14-4 x^{2}-10 x+6}{(2 x+7)^{2} \sqrt{2 x-1}}=\frac{2 x^{2}+15 x+20}{(2 x+7)^{2} \sqrt{2 x-1}}
\end{aligned}
$$

Problem Kuznetsov Differentiation 6-29