Condition of the problem

Find the derivative of the specified order.

$y=\left(1+x^{2}\right) \operatorname{arctg} x, y^{\prime \prime \prime}=?$
## Solution

$y^{\prime}=\left(\left(1+x^{2}\right) \operatorname{arctg} x\right)^{\prime}=2 x \cdot \operatorname{arctg} x+\left(1+x^{2}\right) \cdot \frac{1}{1+x^{2}}=$

$=2 x \cdot \operatorname{arctg} x+1$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=(2 x \cdot \operatorname{arctg} x+1)^{\prime}=2 \operatorname{arctg} x+2 x \cdot \frac{1}{1+x^{2}}=$

$=2 \operatorname{arctg} x+\frac{2 x}{1+x^{2}}$

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

$=\frac{2\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}+\frac{2+2 x^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}=\frac{4+4 x^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}=\frac{4}{\left(1+x^{2}\right)^{2}}$

Problem Kuznetsov Differentiation $19-10$