## Task Condition

Find the derivative of the specified order.

$y=\left(2 x^{2}-7\right) \ln (x-1), y^{V}=?$
## Solution

$y^{\prime}=\left(\left(2 x^{2}-7\right) \ln (x-1)\right)^{\prime}=4 x \cdot \ln (x-1)+\left(2 x^{2}-7\right) \cdot \frac{1}{x-1}=$

$=4 x \cdot \ln (x-1)+\frac{2 x^{2}-7}{x-1}$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(4 x \cdot \ln (x-1)+\frac{2 x^{2}-7}{x-1}\right)^{\prime}=$

$=4 \ln (x-1)+4 x \cdot \frac{1}{x-1}+\frac{4 x \cdot(x-1)-\left(2 x^{2}-7\right) \cdot 1}{(x-1)^{2}}=$

$=4 \ln (x-1)+\frac{4 x^{2}-4 x}{(x-1)^{2}}+\frac{4 x^{2}-4 x-2 x^{2}+7}{(x-1)^{2}}=$

$=4 \ln (x-1)+\frac{6 x^{2}-8 x+7}{(x-1)^{2}}$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(4 \ln (x-1)+\frac{6 x^{2}-8 x+7}{(x-1)^{2}}\right)^{\prime}=$

$=4 \cdot \frac{1}{x-1}+\frac{(12 x-8) \cdot(x-1)^{2}-\left(6 x^{2}-8 x+7\right) \cdot 2(x-1)}{(x-1)^{4}}=$

$=\frac{4(x-1)^{2}}{(x-1)^{3}}+\frac{(12 x-8) \cdot(x-1)-\left(6 x^{2}-8 x+7\right) \cdot 2}{(x-1)^{3}}=$

$=\frac{4 x^{2}-8 x+4+12 x^{2}-12 x-8 x+8-12 x^{2}+16 x-14}{(x-1)^{3}}=$

$=\frac{4 x^{2}-12 x-2}{(x-1)^{3}}$

$y^{(4)}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(\frac{4 x^{2}-12 x-2}{(x-1)^{3}}\right)^{\prime}=$

$$
\begin{aligned}
& =\frac{(8 x-12) \cdot(x-1)^{3}-\left(4 x^{2}-12 x-2\right) \cdot 3(x-1)^{2}}{(x-1)^{6}}= \\
& =\frac{(8 x-12) \cdot(x-1)-\left(4 x^{2}-12 x-2\right) \cdot 3}{(x-1)^{4}}= \\
& =\frac{8 x^{2}-12 x-8 x+12-12 x^{2}+36 x+6}{(x-1)^{4}}=\frac{-4 x^{2}+16 x+18}{(x-1)^{4}} \\
& y^{(5)}=\left(y^{(4)}\right)^{\prime}=\left(\frac{-4 x^{2}+16 x+18}{(x-1)^{4}}\right)^{\prime}= \\
& =\frac{(-8 x+16) \cdot(x-1)^{4}-\left(-4 x^{2}+16 x+18\right) \cdot 4(x-1)^{3}}{(x-1)^{8}}= \\
& =\frac{(-8 x+16) \cdot(x-1)-\left(-4 x^{2}+16 x+18\right) \cdot 4}{(x-1)^{5}}= \\
& =\frac{-8 x^{2}+16 x+8 x-16+16 x^{2}-64 x-72}{(x-1)^{5}}= \\
& =\frac{8 x^{2}-40 x-88}{(x-1)^{5}}=\frac{8\left(x^{2}-5 x-11\right)}{(x-1)^{5}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 19-1