## Task Condition

Find the derivative of the specified order.

$$
y=\frac{\ln (2 x+5)}{2 x+5}, y^{\prime \prime \prime}=?
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{\ln (2 x+5)}{2 x+5}\right)^{\prime}=\frac{\frac{1}{2 x+5} \cdot 2 \cdot(2 x+5)-\ln (2 x+5) \cdot 2}{(2 x+5)^{2}}= \\
& =\frac{2-2 \ln (2 x+5)}{(2 x+5)^{2}}=2 \cdot \frac{1-\ln (2 x+5)}{(2 x+5)^{2}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(2 \cdot \frac{1-\ln (2 x+5)}{(2 x+5)^{2}}\right)^{\prime}= \\
& =2 \cdot \frac{-\frac{1}{2 x+5} \cdot 2 \cdot(2 x+5)^{2}-(1-\ln (2 x+5)) \cdot 2(2 x+5) \cdot 2}{(2 x+5)^{4}}= \\
& =2 \cdot \frac{-2-4(1-\ln (2 x+5))}{(2 x+5)^{3}}=-4 \cdot \frac{1+2(1-\ln (2 x+5))}{(2 x+5)^{3}}= \\
& =-4 \cdot \frac{3-2 \ln (2 x+5)}{(2 x+5)^{3}} \\
& y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-4 \cdot \frac{3-2 \ln (2 x+5)}{(2 x+5)^{3}}\right)^{\prime}= \\
& =-4 \cdot \frac{\frac{-2}{2 x+5} \cdot 2 \cdot(2 x+5)^{3}-(3-2 \ln (2 x+5)) \cdot 3(2 x+5)^{2} \cdot 2}{(2 x+5)^{6}}= \\
& =-8 \cdot \frac{-2-3(3-2 \ln (2 x+5))}{(2 x+5)^{4}}=-8 \cdot \frac{-11+6 \ln (2 x+5)}{(2 x+5)^{4}}= \\
& =\frac{88-48 \ln (2 x+5)}{(2 x+5)^{4}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 19-21