## Task Condition

Find the $n$-th order derivative.

$y=\lg (3 x+1)$
## Solution

$$
\begin{aligned}
& y=\lg (3 x+1) \\
& y^{\prime}=(\lg (3 x+1))^{\prime}=\frac{3}{(3 x+1) \ln 10}=\frac{3}{\ln 10} \cdot(3 x+1)^{-1} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{3}{\ln 10} \cdot(3 x+1)^{-1}\right)^{\prime}=-\frac{3^{2}}{\ln 10} \cdot(3 x+1)^{-2} \\
& y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-\frac{3^{2}}{\ln 10} \cdot(3 x+1)^{-2}\right)^{\prime}=\frac{2 \cdot 3^{3}}{\ln 10} \cdot(3 x+1)^{-3} \\
& y^{(4)}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(\frac{2 \cdot 3^{3}}{\ln 10} \cdot(3 x+1)^{-3}\right)^{\prime}=-\frac{2 \cdot 3 \cdot 3^{4}}{\ln 10} \cdot(3 x+1)^{-4}
\end{aligned}
$$

Obviously,

$$
y^{(n)}=\frac{(-1)^{n-1} \cdot(n-1)!\cdot 3^{n}}{\ln 10} \cdot(3 x+1)^{-n}=\frac{(-1)^{n-1} \cdot(n-1)!\cdot 3^{n}}{\ln 10 \cdot(3 x+1)^{n}}
$$

## Problem Kuznetsov Differentiation 18-15