## Task Condition

Find the $n$-th order derivative.

$$
y=\lg (5 x+2)
$$
## Solution

$y^{\prime}=\frac{1}{(5 x+2) \cdot \ln 10} \cdot 5=\frac{5}{(5 x+2) \cdot \ln 10}$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{5}{(5 x+2) \cdot \ln 10}\right)^{\prime}=\frac{5}{\ln 10} \cdot\left(-\frac{1}{(5 x+2)^{2}} \cdot 5\right)=$

$=-\frac{5^{2}}{\ln 10} \cdot \frac{1}{(5 x+2)^{2}}$
$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-\frac{5^{2}}{\ln 10} \cdot \frac{1}{(5 x+2)^{2}}\right)^{\prime}=-\frac{5^{2}}{\ln 10} \cdot \frac{-2}{(5 x+2)^{3}} \cdot 5=$ $=\frac{2 \cdot 5^{3}}{\ln 10} \cdot \frac{1}{(5 x+2)^{3}}$

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

Obviously,

$y^{(n)}=\frac{(-1)^{n-1} \cdot(n-1)!\cdot 5^{n}}{\ln 10 \cdot(5 x+2)^{n}}$

## Problem Kuznetsov Differentiation 18-5