## Task Condition

Find the derivative of the specified order.

$y=e^{-x} \cdot(\cos 2 x-3 \sin 2 x), y^{IV}=?$
## Solution

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

$=-e^{-x} \cdot(\cos 2 x-3 \sin 2 x)+e^{-x} \cdot(-2 \sin 2 x-6 \cos 2 x)=$

$=-e^{-x} \cdot(\cos 2 x-3 \sin 2 x+2 \sin 2 x+6 \cos 2 x)=$

$=-e^{-x} \cdot(7 \cos 2 x-\sin 2 x)$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(-e^{-x} \cdot(7 \cos 2 x-\sin 2 x)\right)^{\prime}=$

$=e^{-x} \cdot(7 \cos 2 x-\sin 2 x)-e^{-x} \cdot(-14 \sin 2 x-2 \cos 2 x)=$

$=e^{-x} \cdot(7 \cos 2 x-\sin 2 x+14 \sin 2 x+2 \cos 2 x)=$

$=e^{-x} \cdot(9 \cos 2 x+13 \sin 2 x)$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(e^{-x} \cdot(9 \cos 2 x+13 \sin 2 x)\right)^{\prime}=$

$=-e^{-x} \cdot(9 \cos 2 x+13 \sin 2 x)+e^{-x} \cdot(-18 \sin 2 x+26 \cos 2 x)=$

$=-e^{-x} \cdot(9 \cos 2 x+13 \sin 2 x+18 \sin 2 x-26 \cos 2 x)=$

$=-e^{-x} \cdot(31 \sin 2 x-17 \cos 2 x)$

$y^{I V}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(-e^{-x} \cdot(31 \sin 2 x-17 \cos 2 x)\right)^{\prime}=$

$=e^{-x} \cdot(31 \sin 2 x-17 \cos 2 x)-e^{-x} \cdot(62 \cos 2 x+34 \sin 2 x)=$

$=e^{-x} \cdot(31 \sin 2 x-17 \cos 2 x-62 \cos 2 x-34 \sin 2 x)=$

$=-e^{-x} \cdot(79 \cos 2 x+3 \sin 2 x)$

## Problem Kuznetsov Differentiation 19-28