## Task Condition

Find the $n$-th order derivative.

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

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

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

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

$y^{(I V)}=\left(y^{\prime \prime \prime}\right)^{\prime}=(-8 \cos 2 x+\sin (x+1))^{\prime}=16 \sin 2 x+\cos (x+1)$

$\cdots$

Obviously,

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

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

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

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

Or written in one expression:

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

## Problem Kuznetsov Differentiation 18-2