## Task Condition

Find the derivative of the specified order.

$$
y=x \ln (1-3 x), y^{(I V)}=?
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=(x \ln (1-3 x))^{\prime}=\ln (1-3 x)+\frac{x}{1-3 x} \cdot(-3)=\ln (1-3 x)-\frac{3 x}{1-3 x} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\ln (1-3 x)-\frac{3 x}{1-3 x}\right)^{\prime}=\frac{1}{1-3 x} \cdot(-3)-\frac{3 \cdot(1-3 x)-3 x \cdot(-3)}{(1-3 x)^{2}}= \\
& =-\frac{3}{1-3 x}-\frac{3}{(1-3 x)^{2}}=-3 \cdot\left(\frac{1}{1-3 x}+\frac{1}{(1-3 x)^{2}}\right) \\
& y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-3 \cdot\left(\frac{1}{1-3 x}+\frac{1}{(1-3 x)^{2}}\right)\right)^{\prime}=-3 \cdot\left(\frac{-1}{(1-3 x)^{2}} \cdot(-3)+\frac{-2}{(1-3 x)^{3}} \cdot(-3)\right)= \\
& =-9 \cdot\left(\frac{1}{(1-3 x)^{2}}+\frac{2}{(1-3 x)^{3}}\right) \\
& y^{(I V)}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(-9 \cdot\left(\frac{1}{(1-3 x)^{2}}+\frac{2}{(1-3 x)^{3}}\right)\right)^{\prime}= \\
& =-9 \cdot\left(\frac{-2}{(1-3 x)^{3}} \cdot(-3)+\frac{-3 \cdot 2}{(1-3 x)^{4}} \cdot(-3)\right)=-27 \cdot\left(\frac{2}{(1-3 x)^{3}}+\frac{6}{(1-3 x)^{4}}\right)= \\
& =-54 \cdot\left(\frac{1-3 x}{(1-3 x)^{4}}+\frac{3}{(1-3 x)^{4}}\right)=-\frac{54(4-3 x)}{(1-3 x)^{4}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation $19-24$