## Task Condition

Find the derivative of the specified order.

$$
y=(5 x-1) \ln ^{2} x, y^{\prime \prime \prime}=?
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left((5 x-1) \ln ^{2} x\right)^{\prime}=5 \ln ^{2} x+(5 x-1) \cdot 2 \cdot \ln x \cdot \frac{1}{x}= \\
& =5 \ln ^{2} x+2 \cdot \frac{(5 x-1) \cdot \ln x}{x} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(5 \ln ^{2} x+2 \cdot \frac{(5 x-1) \cdot \ln x}{x}\right)^{\prime}= \\
& =10 \ln x \cdot \frac{1}{x}+2 \cdot \frac{\left(5 \cdot \ln x+(5 x-1) \cdot \frac{1}{x}\right) \cdot x-(5 x-1) \cdot \ln x \cdot 1}{x^{2}}= \\
& =\frac{10 x \cdot \ln x}{x^{2}}+2 \cdot \frac{5 x \cdot \ln x+5 x-1-5 x \cdot \ln x+\ln x}{x^{2}}= \\
& =\frac{10 x \cdot \ln x+10 x-2+2 \ln x}{x^{2}}=2 \cdot \frac{5 x \cdot(\ln x+1)+\ln x-1}{x^{2}} \\
& y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(2 \cdot \frac{5 x \cdot(\ln x+1)+\ln x-1}{x^{2}}\right)^{\prime}= \\
& =2 \cdot \frac{\left(5 \cdot(\ln x+1)+5 x \cdot \frac{1}{x}+\frac{1}{x}\right) \cdot x^{2}-(5 x \cdot(\ln x+1)+\ln x-1) \cdot 2 x}{x^{4}}= \\
& =2 \cdot \frac{5 x \cdot(\ln x+1)+5 x+1-2(5 x \cdot(\ln x+1)+\ln x-1)}{x^{3}}= \\
& =2 \cdot \frac{5 x \cdot \ln x+5 x+5 x+1-10 x \cdot \ln x-10 x-2 \ln x+2}{x^{3}}= \\
& =2 \cdot \frac{3-(5 x+2) \ln x}{x^{3}}=\frac{6-2(5 x+2) \ln x}{x^{3}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 19-29