Condition of the problem

Find the derivative.

$y=\ln \left(\frac{\ln x}{\sin \frac{1}{x}}\right)$
## Solution

$y^{\prime}=\left(\ln \left(\frac{\ln x}{\sin \frac{1}{x}}\right)\right)^{\prime}=\frac{\sin \frac{1}{x}}{\ln x} \cdot\left(\frac{\ln x}{\sin \frac{1}{x}}\right)^{\prime}=$

$=\frac{\sin \frac{1}{x}}{\ln x} \cdot \frac{\frac{1}{x} \cdot \sin \frac{1}{x}-\ln x \cdot \cos \frac{1}{x}}{\sin ^{2} \frac{1}{x}}=\frac{\sin \frac{1}{x}-x \cdot \ln x \cdot \cos \frac{1}{x}}{x \cdot \ln x \cdot \sin \frac{1}{x}}$

## Problem Kuznetsov Differentiation 8-28