## Task Condition

Find the $n$-th order derivative.

$y=\frac{x}{9(4 x+9)}$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{x}{9(4 x+9)}\right)^{\prime}=\frac{1 \cdot(4 x+9)-x \cdot 4}{9(4 x+9)^{2}}=\frac{4 x+9-4 x}{9(4 x+9)^{2}}=\frac{1}{(4 x+9)^{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1}{(4 x+9)^{2}}\right)^{\prime}=\frac{-2}{(4 x+9)^{3}} \cdot 4=\frac{-2 \cdot 4}{(4 x+9)^{3}}
\end{aligned}
$$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(\frac{-2 \cdot 4}{(4 x+9)^{3}}\right)^{\prime}=\frac{-2 \cdot 4 \cdot(-3)}{(4 x+9)^{4}} \cdot 4=\frac{2 \cdot 3 \cdot 4^{2}}{(4 x+9)^{4}}$

$\cdots$

Obviously,

$$
y^{(n)}=\frac{(-1)^{n-1} \cdot n!\cdot 4^{n-1}}{(4 x+9)^{n+1}}
$$

## Problem Kuznetsov Differentiation $18-17$