## Task Condition

Find the derivative of the specified order.

$$
y=\frac{\ln (3+x)}{3+x}, y^{\prime \prime \prime}=?
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{\ln (3+x)}{3+x}\right)^{\prime}=\frac{\frac{1}{3+x} \cdot(3+x)-\ln (3+x) \cdot 1}{(3+x)^{2}}=\frac{1-\ln (3+x)}{(3+x)^{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1-\ln (3+x)}{(3+x)^{2}}\right)^{\prime}=\frac{-\frac{1}{3+x} \cdot(3+x)^{2}-(1-\ln (3+x)) \cdot 2(3+x)}{(3+x)^{4}}= \\
& =\frac{-1-(1-\ln (3+x)) \cdot 2}{(3+x)^{3}}=\frac{-1-2+2 \ln (3+x)}{(3+x)^{3}}=\frac{-3+2 \ln (3+x)}{(3+x)^{3}} \\
& y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(\frac{-3+2 \ln (3+x)}{(3+x)^{3}}\right)^{\prime}= \\
& =\frac{\frac{2}{3+x} \cdot(3+x)^{3}-(-3+2 \ln (3+x)) \cdot 3(3+x)^{2}}{(3+x)^{6}}= \\
& =\frac{2-(-3+2 \ln (3+x)) \cdot 3}{(3+x)^{4}}=\frac{2+9-6 \ln (3+x)}{(3+x)^{4}}=\frac{11-6 \ln (3+x)}{(3+x)^{4}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 19-14