Condition of the problem

Find the $n$-th order derivative.

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

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

$y^{\prime}=\left(\frac{2 x+5}{13(3 x+1)}\right)^{\prime}=\frac{2 \cdot(3 x+1)-(2 x+5) \cdot 3}{13(3 x+1)^{2}}=$

$=\frac{6 x+2-6 x-15}{13(3 x+1)^{2}}=-\frac{13}{13(3 x+1)^{2}}=-\frac{1}{(3 x+1)^{2}}$

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

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

...

It is obvious that

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

## Problem Kuznetsov Differentiation $18-10$