## Task Condition

Find the derivative of the specified order.

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

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

$=2 x \cdot \ln (x-3)+\frac{x^{2}+3}{x-3}$

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

$=2 \ln (x-3)+2 x \cdot \frac{1}{x-3}+\frac{2 x \cdot(x-3)-\left(x^{2}+3\right) \cdot 1}{(x-3)^{2}}=$

$=2 \ln (x-3)+\frac{2 x(x-3)}{(x-3)^{2}}+\frac{2 x^{2}-6 x-x^{2}-3}{(x-3)^{2}}=$

$=2 \ln (x-3)+\frac{3 x^{2}-12 x-3}{(x-3)^{2}}$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(2 \ln (x-3)+\frac{3 x^{2}-12 x-3}{(x-3)^{2}}\right)^{\prime}=$

$=\frac{2}{x-3}+\frac{(6 x-12) \cdot(x-3)^{2}-\left(3 x^{2}-12 x-3\right) \cdot 2(x-3)}{(x-3)^{4}}=$

$=\frac{2(x-3)^{2}}{(x-3)^{3}}+\frac{(6 x-12) \cdot(x-3)-\left(3 x^{2}-12 x-3\right) \cdot 2}{(x-3)^{3}}=$

$=\frac{2 x^{2}-12 x+18+6 x^{2}-18 x-12 x+36-6 x^{2}+24 x+6}{(x-3)^{3}}=$

$=\frac{2 x^{2}-18 x+60}{(x-3)^{3}}$

$y^{I V}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(\frac{2 x^{2}-18 x+60}{(x-3)^{3}}\right)^{\prime}=$

$$
\begin{aligned}
& =\frac{(4 x-18) \cdot(x-3)^{3}-\left(2 x^{2}-18 x+60\right) \cdot 3(x-3)^{2}}{(x-3)^{6}}= \\
& =\frac{(4 x-18) \cdot(x-3)-\left(2 x^{2}-18 x+60\right) \cdot 3}{(x-3)^{4}}= \\
& =\frac{4 x^{2}-12 x-18 x+54-6 x^{2}+54 x-180}{(x-3)^{4}}=\frac{-2 x^{2}+24 x-126}{(x-3)^{4}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 19-16