Condition of the problem

Find the derivative of the specified order.

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

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

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

$=\frac{1}{4} \cdot e^{\frac{x}{2}} \cdot \sin 2 x+e^{\frac{x}{2}} \cdot \cos 2 x+e^{\frac{x}{2}} \cdot \cos 2 x-4 e^{\frac{x}{2}} \cdot \sin 2 x=$

$=-\frac{15}{4} \cdot e^{\frac{x}{2}} \cdot \sin 2 x+2 e^{\frac{x}{2}} \cdot \cos 2 x$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-\frac{15}{4} \cdot e^{\frac{x}{2}} \cdot \sin 2 x+2 e^{\frac{x}{2}} \cdot \cos 2 x\right)^{\prime}=$

$=-\frac{15}{8} \cdot e^{\frac{x}{2}} \cdot \sin 2 x-\frac{15}{2} \cdot e^{\frac{x}{2}} \cdot \cos 2 x+e^{\frac{x}{2}} \cdot \cos 2 x-4 e^{\frac{x}{2}} \cdot \sin 2 x=$

$=-\frac{47}{8} \cdot e^{\frac{x}{2}} \cdot \sin 2 x-\frac{13}{2} \cdot e^{\frac{x}{2}} \cdot \cos 2 x$

$y^{I V}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(-\frac{47}{8} \cdot e^{\frac{x}{2}} \cdot \sin 2 x-\frac{13}{2} \cdot e^{\frac{x}{2}} \cdot \cos 2 x\right)^{\prime}=$

$=-\frac{47}{16} \cdot e^{\frac{x}{2}} \cdot \sin 2 x-\frac{47}{4} \cdot e^{\frac{x}{2}} \cdot \cos 2 x-\frac{13}{4} \cdot e^{\frac{x}{2}} \cdot \cos 2 x+13 \cdot e^{\frac{x}{2}} \cdot \sin 2 x=$ $=\frac{161}{16} \cdot e^{\frac{x}{2}} \cdot \sin 2 x-\frac{60}{4} \cdot e^{\frac{x}{2}} \cdot \cos 2 x=\frac{161}{16} \cdot e^{\frac{x}{2}} \cdot \sin 2 x-15 \cdot e^{\frac{x}{2}} \cdot \cos 2 x$

## Problem Kuznetsov Differentiation 19-22