## Task Condition

Find the derivative.

$y=\left(2 x^{2}+6 x+5\right) \operatorname{arctg} \frac{x+1}{x+2}-x$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\left(2 x^{2}+6 x+5\right) \operatorname{arctg} \frac{x+1}{x+2}-x\right)^{\prime}= \\
& =\left(2 x^{2}+6 x+5\right)^{\prime} \operatorname{arctg} \frac{x+1}{x+2}+\left(2 x^{2}+6 x+5\right)\left(\operatorname{arctg} \frac{x+1}{x+2}\right)^{\prime}-1= \\
& =(4 x+6) \operatorname{arctg} \frac{x+1}{x+2}+\left(2 x^{2}+6 x+5\right) \frac{1}{1+\left(\frac{x+1}{x+2}\right)^{2}} \cdot\left(\frac{x+1}{x+2}\right)^{\prime}-1= \\
& =(4 x+6) \operatorname{arctg} \frac{x+1}{x+2}+\frac{2 x^{2}+6 x+5}{1+\left(\frac{x+1}{x+2}\right)^{2}} \cdot \frac{1 \cdot(x+2)-(x+1) \cdot 1}{(x+2)^{2}}-1= \\
& =(4 x+6) \operatorname{arctg} \frac{x+1}{x+2}+\frac{\left(2 x^{2}+6 x+5\right)(x+2-x-1)}{(x+2)^{2}+(x+1)^{2}}-1= \\
& =(4 x+6) \operatorname{arctg} \frac{x+1}{x+2}+\frac{2 x^{2}+6 x+5}{2 x^{2}+6 x+5}-1= \\
& =(4 x+6) \operatorname{arctg} \frac{x+1}{x+2}+1-1=(4 x+6) \operatorname{arctg} \frac{x+1}{x+2}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 10-26