## Problem Statement

Find the $n$-th order derivative.

$y=\frac{4 x+7}{2 x+3}$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{4 x+7}{2 x+3}\right)^{\prime}=\frac{4 \cdot(2 x+3)-(4 x+7) \cdot 2}{(2 x+3)^{2}}=\frac{8 x+12-8 x-14}{(2 x+3)^{2}}=-\frac{2}{(2 x+3)^{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(-\frac{2}{(2 x+3)^{2}}\right)^{\prime}=-\frac{2 \cdot(-2)}{(2 x+3)^{3}} \cdot 2=\frac{2^{2} \cdot 1 \cdot 2}{(2 x+3)^{3}} \\
& y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(\frac{2^{2} \cdot 1 \cdot 2}{(2 x+3)^{3}}\right)^{\prime}=\frac{2^{2} \cdot 1 \cdot 2 \cdot(-3)}{(2 x+3)^{4}} \cdot 2=-\frac{2^{3} \cdot 1 \cdot 2 \cdot 3}{(2 x+3)^{4}}
\end{aligned}
$$

$\cdots$

Obviously,

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

## Problem Kuznetsov Differentiation 18-4