## Task Condition

Find the $n$-th order derivative.

$$
y=\frac{x}{x+1}
$$
## Solution

To derive the formula of the $\mathrm{N}$-th order, we first find several derivatives and then establish a general rule suitable for calculating the derivative of any order:

$$
\begin{aligned}
y^{\prime} & =\frac{x+1-x}{(x+1)^{2}}=\frac{1}{(x+1)^{2}} \\
y^{\prime \prime} & =-\frac{1 \cdot 2(x+1)}{(x+1)^{4}}=-\frac{1 \cdot 2}{(x+1)^{3}} \\
y^{\prime \prime \prime} & =\frac{1 \cdot 2 \cdot 3(x+1)^{2}}{(x+1)^{6}}=\frac{1 \cdot 2 \cdot 3}{(x+1)^{4}} \\
y^{(n)} & =(-1)^{n+1} \cdot \frac{n!}{(x+1)^{n+1}}
\end{aligned}
$$

We will prove the obtained formula by mathematical induction:

$$
\begin{aligned}
& y^{(k)}=(-1)^{k+1} \cdot \frac{k!}{(x+1)^{k+1}} \\
& y^{(k+1)}={ }^{?}(-1)^{k+2} \cdot \frac{(k+1)!}{(x+1)^{k+2}} \\
& y^{(k+1)}=\left(y^{(k)}\right)^{\prime}=(-1)^{k+1} \cdot k!\cdot\left((x+1)^{-(k+1)}\right)^{\prime}=(-1)^{k+1} \cdot k!\cdot(-(k+1)) \cdot(x+1)^{-(k+1)-1}= \\
& =(-1)(-1)^{k+1} \cdot k!\cdot(k+1) \cdot \frac{1}{(x+1)^{k+2}}=(-1)^{k+2} \cdot \frac{(k+1)!}{(x+1)^{k+2}}
\end{aligned}
$$

This completes the proof!

## Problem Kuznetsov Differentiation 18-27