## Task Condition

Find the derivative of the specified order.

$$
y=\left(2 x^{3}+1\right) \cos x, y^{V}=?
$$
## Solution

$y^{\prime}=\left(\left(2 x^{3}+1\right) \cos x\right)^{\prime}=6 x^{2} \cdot \cos x-\left(2 x^{3}+1\right) \sin x_{.}$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(6 x^{2} \cdot \cos x-\left(2 x^{3}+1\right) \sin x\right)^{\prime}=$

$=12 x \cdot \cos x-6 x^{2} \cdot \sin x-6 x^{2} \cdot \sin x-\left(2 x^{3}+1\right) \cos x=$

$=-12 x^{2} \cdot \sin x-\left(2 x^{3}-12 x+1\right) \cos x$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(-12 x^{2} \cdot \sin x-\left(2 x^{3}-12 x+1\right) \cos x\right)^{\prime}=$

$=-24 x \cdot \sin x-12 x^{2} \cdot \cos x-\left(6 x^{2}-12\right) \cos x+\left(2 x^{3}-12 x+1\right) \sin x=$

$=-\left(18 x^{2}-12\right) \cos x+\left(2 x^{3}-36 x+1\right) \sin x$
$y^{(IV)}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(-\left(18 x^{2}-12\right) \cos x+\left(2 x^{3}-36 x+1\right) \sin x\right)^{\prime}=$

$=-36 x \cdot \cos x+\left(18 x^{2}-12\right) \sin x+\left(6 x^{2}-36\right) \sin x+\left(2 x^{3}-36 x+1\right) \cos x=$ $=\left(24 x^{2}-48\right) \sin x+\left(2 x^{3}-72 x+1\right) \cos x$

$y^{(V)}=\left(\left(24 x^{2}-48\right) \sin x+\left(2 x^{3}-72 x+1\right) \cos x\right)^{\prime}=$

$=48 x^{2} \cdot \sin x+\left(24 x^{2}-48\right) \cos x+\left(6 x^{2}-72\right) \cos x-\left(2 x^{3}-72 x+1\right) \sin x=$ $=\left(30 x^{2}-120\right) \cos x-\left(2 x^{3}-120 x+1\right) \sin x$

## Problem Kuznetsov Differentiation 19-15