## Task Condition

Find the derivative of the specified order.

$$
y=\frac{\ln (x-1)}{\sqrt{x-1}}, y^{\prime \prime \prime}=?
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{\ln (x-1)}{\sqrt{x-1}}\right)^{\prime}=\frac{\frac{1}{x-1} \cdot \sqrt{x-1}-\ln (x-1) \cdot \frac{1}{2 \sqrt{x-1}}}{x-1}=\frac{2-\ln (x-1)}{2 \sqrt{(x-1)^{3}}} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{2-\ln (x-1)}{2 \sqrt{(x-1)^{3}}}\right)^{\prime}=\frac{-\frac{1}{x-1} \cdot \sqrt{(x-1)^{3}}-(2-\ln (x-1)) \cdot \frac{3}{2} \cdot \sqrt{x-1}}{2(x-1)^{3}}= \\
& =\frac{-2-3(2-\ln (x-1))}{4 \sqrt{(x-1)^{5}}}=\frac{-8+3 \ln (x-1)}{4 \sqrt{(x-1)^{5}}} \\
& y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(\frac{-8+3 \ln (x-1)}{4 \sqrt{(x-1)^{5}}}\right)^{\prime}=\frac{\frac{3}{x-1} \cdot \sqrt{(x-1)^{5}}-(-8+3 \ln (x-1)) \cdot \frac{5}{2} \cdot \sqrt{(x-1)^{3}}}{4(x-1)^{5}}= \\
& =\frac{6-5(-8+3 \ln (x-1))}{8 \sqrt{(x-1)^{7}}}=\frac{46-15 \ln (x-1)}{8 \sqrt{(x-1)^{7}}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 19-4