## Problem Statement

Find the $n$-th order derivative.

$y=\frac{4}{x}$
## Solution

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

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

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

$\cdots$

Obviously,

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

## Problem Kuznetsov Differentiation 18-19