## Task Condition

Find the $n$-th order derivative.

$y=\sin (x+1)+\cos 2 x$
## Solution

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

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=(\cos (x+1)-2 \sin 2 x)^{\prime}=-\sin (x+1)-4 \cos 2 x$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=(\sin (x+1)-4 \cos 2 x)^{\prime}=-\cos (x+1)+8 \sin 2 x$

$y^{(I V)}=\left(y^{\prime \prime \prime}\right)^{\prime}=(-\cos (x+1)+8 \sin 2 x)^{\prime}=\sin (x+1)+2^{4} \cdot \cos 2 x$

$\cdots$

Obviously,

$y^{(4 n)}=\sin (x+1)+2^{4 n} \cdot \cos 2 x$

$y^{(4 n+1)}=\cos (x+1)-2^{4 n+1} \cdot \sin 2 x$

$y^{(4 n+2)}=-\sin (x+1)-2^{4 n+2} \cdot \cos 2 x$

$y^{(4 n+3)}=-\cos (x+1)+2^{4 n+3} \cdot \sin 2 x$

Or written in one expression:

$y^{(n)}=\sin \left(\frac{3 \pi}{2} \cdot n+x+1\right)+2^{n} \cdot \cos \left(\frac{3 \pi}{2} \cdot n+2 x\right)$

## Problem Kuznetsov Differentiation 18-12