## problem statement

Find the $n$-th order derivative.

$y=\sin (3 x+1)+\cos 5 x$
## Solution

$y^{\prime}=(\sin (3 x+1)+\cos 5 x)^{\prime}=3 \cos (3 x+1)-5 \sin 5 x$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=(3 \cos (3 x+1)-5 \sin 5 x)^{\prime}=-3^{2} \cdot \sin (3 x+1)-5^{2} \cdot \cos 5 x$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-3^{2} \cdot \sin (3 x+1)-5^{2} \cdot \cos 5 x\right)^{\prime}=-3^{3} \cdot \cos (3 x+1)+5^{3} \cdot \sin 5 x$

$y^{(I V)}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(-3^{3} \cdot \cos (3 x+1)+5^{3} \cdot \sin 5 x\right)^{\prime}=3^{4} \cdot \sin (3 x+1)+5^{4} \cdot \cos 5 x$

$\cdots$

Obviously,

$y^{(4 n)}=3^{4 n} \cdot \sin (3 x+1)+5^{4 n} \cdot \cos 5 x$

$y^{(4 n+1)}=3^{4 n+1} \cdot \cos (3 x+1)-5^{4 n+1} \cdot \sin 5 x$

$y^{(4 n+2)}=-3^{4 n+2} \cdot \sin (3 x+1)-5^{4 n+2} \cdot \cos 5 x$

$y^{(4 n+3)}=-3^{4 n+3} \cdot \cos (3 x+1)+5^{4 n+3} \cdot \sin 5 x$

Or written in one expression:

$y^{(n)}=3^{n} \cdot \sin \left(\frac{3 \pi}{2} \cdot n+3 x+1\right)+5^{n} \cdot \cos \left(\frac{3 \pi}{2} \cdot n+5 x\right)$

## Problem Kuznetsov Differentiation 18-22