## Task Condition

Find the derivative of the specified order.

$$
y=\frac{1}{x} \cdot \sin 2 x, y^{\prime \prime \prime}=?
$$
## Solution

$y^{\prime}=\left(\frac{1}{x} \cdot \sin 2 x\right)^{\prime}=-\frac{1}{x^{2}} \cdot \sin 2 x+\frac{2}{x} \cdot \cos 2 x$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(-\frac{1}{x^{2}} \cdot \sin 2 x+\frac{2}{x} \cdot \cos 2 x\right)^{\prime}=$

$=\frac{2}{x^{3}} \cdot \sin 2 x-\frac{2}{x^{2}} \cdot \cos 2 x-\frac{2}{x^{2}} \cdot \cos 2 x-\frac{4}{x} \cdot \sin 2 x=$

$=\frac{2}{x^{3}} \cdot \sin 2 x-\frac{4}{x^{2}} \cdot \cos 2 x-\frac{4}{x} \cdot \sin 2 x$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(\frac{2}{x^{3}} \cdot \sin 2 x-\frac{4}{x^{2}} \cdot \cos 2 x-\frac{4}{x} \cdot \sin 2 x\right)^{\prime}=$

$=-\frac{6}{x^{4}} \cdot \sin 2 x+\frac{4}{x^{3}} \cdot \cos 2 x+\frac{8}{x^{3}} \cdot \cos 2 x+\frac{8}{x^{2}} \cdot \sin 2 x+\frac{4}{x^{2}} \cdot \sin 2 x-\frac{8}{x} \cdot \cos 2 x=$ $=-\frac{6}{x^{4}} \cdot \sin 2 x+\frac{12}{x^{3}} \cdot \cos 2 x+\frac{12}{x^{2}} \cdot \sin 2 x-\frac{8}{x} \cdot \cos 2 x=$

$=\frac{12 x^{2}-6}{x^{4}} \cdot \sin 2 x+\frac{12-8 x^{2}}{x^{3}} \cdot \cos 2 x$

Problem Kuznetsov Differentiation 19-18