## Task Condition

Find the derivative of the specified order.

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

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

$$
\begin{aligned}
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(-2 x \cdot \ln ^{2} x+\frac{2 \cdot\left(3-x^{2}\right) \cdot \ln x}{x}\right)^{\prime}= \\
& =-2 \cdot \ln ^{2} x-2 x \cdot 2 \cdot \ln x \cdot \frac{1}{x}+\frac{2 \cdot\left(-2 x \cdot \ln x+\left(3-x^{2}\right) \cdot \frac{1}{x}\right) \cdot x-\left(3-x^{2}\right) \cdot \ln x \cdot 1}{x^{2}}= \\
& =-2 \cdot \ln ^{2} x-\frac{4 x^{2} \cdot \ln x}{x^{2}}+\frac{-4 x^{2} \cdot \ln x+2\left(3-x^{2}\right)-\left(3-x^{2}\right) \cdot \ln x}{x^{2}}= \\
& =-2 \cdot \ln ^{2} x+\frac{\left(-7 x^{2}-3\right) \cdot \ln x+6-2 x^{2}}{x^{2}} \\
& y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-2 \cdot \ln ^{2} x+\frac{\left(-7 x^{2}-3\right) \cdot \ln x+6-2 x^{2}}{x^{2}}\right)^{\prime}= \\
& =-4 \cdot \ln x \cdot \frac{1}{x}+\frac{\left(-14 x \cdot \ln x+\left(-7 x^{2}-3\right) \cdot \frac{1}{x}-4 x\right) \cdot x^{2}-\left(\left(-7 x^{2}-3\right) \cdot \ln x+6-2 x^{2}\right) \cdot 2 x}{x^{4}}= \\
& =\frac{-4 \cdot \ln x}{x}+\frac{-14 x^{2} \cdot \ln x-7 x^{2}-3-4 x^{2}+\left(14 x^{2}+6\right) \cdot \ln x-12+4 x^{2}}{x^{3}}= \\
& =\frac{-4 \cdot \ln x}{x}+\frac{6 \ln x-7 x^{2}-15}{x^{3}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 19-2