## Task Condition

Find the derivative of the specified order.

$y=(2 x+3) \ln ^{2} x, y^{\prime \prime \prime}=?$
## Solution

$y^{\prime}=2 \cdot \ln ^{2} x+(2 x+3) \cdot 2 \cdot \ln x \cdot \frac{1}{x}=2 \cdot \ln ^{2} x+\frac{2(2 x+3) \cdot \ln x}{x}$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(2 \cdot \ln ^{2} x+\frac{2(2 x+3) \cdot \ln x}{x}\right)^{\prime}=4 \cdot \ln x \cdot \frac{1}{x}+\left(4 \ln x+\frac{6 \cdot \ln x}{x}\right)^{\prime}=$

$=4 \cdot \ln x \cdot \frac{1}{x}+4 \cdot \frac{1}{x}+6 \cdot \frac{\frac{1}{x} \cdot x-\ln x \cdot 1}{x^{2}}=\frac{4 \cdot \ln x}{x}+\frac{4}{x}+\frac{6}{x^{2}}-\frac{6 \ln x}{x^{2}}$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(\frac{4 \cdot \ln x}{x}+\frac{4}{x}+\frac{6}{x^{2}}-\frac{6 \ln x}{x^{2}}\right)^{\prime}=$

$=4 \frac{\frac{1}{x} \cdot x-\ln x \cdot 1}{x^{2}}-\frac{4}{x^{2}}-2 \cdot \frac{6}{x^{3}}-6 \cdot \frac{\frac{1}{x} \cdot x^{2}-\ln x \cdot 2 x}{x^{4}}=\frac{4-4 \ln x}{x^{2}}-\frac{4}{x^{2}}-\frac{12}{x^{3}}-\frac{6-12 \ln x}{x^{3}}=$ $=\frac{-4 \ln x \cdot x-12-6+12 \ln x}{x^{3}}=\frac{4 \ln x \cdot(3-x)-18}{x^{3}}$

## Problem Kuznetsov Differentiation 19-9