## Task Condition

Find the derivative of the specified order.

$y=(5 x-8) \cdot 2^{-x}, y^{(I V)}=?$
## Solution

$y^{\prime}=\left((5 x-8) \cdot 2^{-x}\right)^{\prime}=5 \cdot 2^{-x}+(5 x-8) \cdot 2^{-x} \cdot(-\ln 2)=2^{-x} \cdot(-5 \ln 2 \cdot x+8 \ln 2+5)$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(2^{-x} \cdot(-5 \ln 2 \cdot x+8 \ln 2+5)\right)^{\prime}=$

$=2^{-x} \cdot(-\ln 2) \cdot(-5 \ln 2 \cdot x+8 \ln 2+5)+2^{-x} \cdot(-5 \ln 2 \cdot)=$

$=2^{-x} \cdot \ln 2 \cdot(5 \ln 2 \cdot x-8 \ln 2-10)$

$y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(2^{-x} \cdot \ln 2 \cdot(5 \ln 2 \cdot x-8 \ln 2-10)\right)^{\prime}=$

$=2^{-x} \cdot(-\ln 2) \cdot \ln 2 \cdot(5 \ln 2 \cdot x-8 \ln 2-10)+2^{-x} \cdot \ln 2 \cdot(5 \ln 2)=$

$=2^{-x} \cdot \ln ^{2} 2 \cdot(-5 \ln 2 \cdot x+8 \ln 2+15)$

$y^{(I V)}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(2^{-x} \cdot \ln ^{2} 2 \cdot(-5 \ln 2 \cdot x+8 \ln 2+15)\right)^{\prime}=$

$=2^{-x} \cdot(-\ln 2) \cdot \ln ^{2} 2 \cdot(-5 \ln 2 \cdot x+8 \ln 2+15)+2^{-x} \cdot \ln ^{2} 2 \cdot(-5 \ln 2)=$

$=2^{-x} \cdot \ln ^{3} 2 \cdot(5 \ln 2 \cdot x-8 \ln 2-20)$

## Problem Kuznetsov Differentiation 19-26