## Task Condition

Find the $n$-th order derivative.

$y=\frac{x}{2(3 x+2)}$
## Solution

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

$\cdots$

Obviously,

$y^{(n)}=\frac{(-1)^{n-1} \cdot n!\cdot 3^{n-1}}{(3 x+2)^{n+1}}$

## Problem Kuznetsov Differentiation 18-7