## Task Condition

Find the derivative of the specified order.

$y=\frac{\log _{3} x}{x^{2}}, y^{IV}=?$
## Solution

$y^{\prime}=\left(\frac{\log _{3} x}{x^{2}}\right)^{\prime}=\left(\frac{\ln x}{x^{2} \cdot \ln 3}\right)^{\prime}=\frac{\frac{1}{x} \cdot x^{2}-\ln x \cdot 2 x}{x^{4} \cdot \ln 3}=\frac{1-2 \ln x}{x^{3} \cdot \ln 3}$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1-2 \ln x}{x^{3} \cdot \ln 3}\right)^{\prime}=\frac{-\frac{2}{x} \cdot x^{3}-(1-2 \ln x) \cdot 3 x^{2}}{x^{6} \cdot \ln 3}=$

$=\frac{-2-3+6 \ln x}{x^{4} \cdot \ln 3}=\frac{-5+6 \ln x}{x^{4} \cdot \ln 3}$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(\frac{-5+6 \ln x}{x^{4} \cdot \ln 3}\right)^{\prime}=\frac{\frac{6}{x} \cdot x^{4}-(-5+6 \ln x) \cdot 4 x^{3}}{x^{8} \cdot \ln 3}=$

$=\frac{6-(-5+6 \ln x) \cdot 4}{x^{5} \cdot \ln 3}=\frac{6+20-24 \ln x}{x^{5} \cdot \ln 3}=\frac{26-24 \ln x}{x^{5} \cdot \ln 3}$

$y^{(I V)}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(\frac{26-24 \ln x}{x^{5} \cdot \ln 3}\right)^{\prime}=\frac{-\frac{24}{x} \cdot x^{5}-(26-24 \ln x) \cdot 5 x^{4}}{x^{10} \cdot \ln 3}=$

$=\frac{-24-(26-24 \ln x) \cdot 5}{x^{6} \cdot \ln 3}=\frac{-24-130+120 \ln x}{x^{6} \cdot \ln 3}=\frac{-154+120 \ln x}{x^{6} \cdot \ln 3}$

## Problem Kuznetsov Differentiation 19-30