## Task Condition

Find the derivative.

$$
y=\ln \left(5 x+\sqrt{25 x^{2}+1}\right)-\sqrt{25 x^{2}+1} \cdot \operatorname{arctg} 5 x
$$
## Solution

$y^{\prime}=\left(\ln \left(5 x+\sqrt{25 x^{2}+1}\right)-\sqrt{25 x^{2}+1} \cdot \operatorname{arctg} 5 x\right)^{\prime}=$

$=\frac{1}{5 x+\sqrt{25 x^{2}+1}} \cdot\left(5+\frac{50 x}{2 \sqrt{25 x^{2}+1}}\right)-\left(\frac{50 x}{2 \sqrt{25 x^{2}+1}} \cdot \operatorname{arctg} 5 x+\sqrt{25 x^{2}+1} \cdot \frac{1}{1+25 x^{2}} \cdot 5\right)=$

$=\frac{1}{5 x+\sqrt{25 x^{2}+1}} \cdot \frac{5 \sqrt{25 x^{2}+1}+25 x}{\sqrt{25 x^{2}+1}}-\left(\frac{25 x}{\sqrt{25 x^{2}+1}} \cdot \operatorname{arctg} 5 x+\frac{5}{\sqrt{25 x^{2}+1}}\right)=$

$=\frac{5}{\sqrt{25 x^{2}+1}}-\frac{25 x \cdot \operatorname{arctg} 5 x}{\sqrt{25 x^{2}+1}}+\frac{5}{\sqrt{25 x^{2}+1}}=-\frac{25 x \cdot \operatorname{arctg} 5 x}{\sqrt{25 x^{2}+1}}$

Problem Kuznetsov Differentiation 13-24