## Problem Statement

Find the $n$-th order derivative.

$y=\frac{5 x+1}{13(2 x+3)}$
## Solution

$$
\begin{aligned}
& y=\frac{5 x+1}{13(2 x+3)} \\
& y^{\prime}=\left(\frac{5 x+1}{13(2 x+3)}\right)^{\prime}=\frac{5 \cdot(2 x+3)-(5 x+1) \cdot 2}{13(2 x+3)^{2}}= \\
& =\frac{10 x+15-10 x-2}{13(2 x+3)^{2}}=\frac{13}{13(2 x+3)^{2}}=\frac{1}{(2 x+3)^{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1}{(2 x+3)^{2}}\right)^{\prime}=\frac{-2}{(2 x+3)^{3}} \cdot 2=-\frac{2 \cdot 2}{(2 x+3)^{3}} \\
& y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-\frac{2 \cdot 2}{(2 x+3)^{3}}\right)^{\prime}=-\frac{2 \cdot(-3) \cdot 2}{(2 x+3)^{4}} \cdot 2=\frac{2 \cdot 3 \cdot 2^{2}}{(2 x+3)^{4}}
\end{aligned}
$$

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

## Problem Kuznetsov Differentiation 18-20