## Task Condition

Find the derivative of the specified order.

$$
y=(4 x+3) \cdot 2^{-x}, y^{V}=?
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left((4 x+3) \cdot 2^{-x}\right)^{\prime}=4 \cdot 2^{-x}+(4 x+3) \cdot 2^{-x} \cdot \ln 2 \cdot(-1)= \\
& =(-\ln 2 \cdot(4 x+3)+4) 2^{-x} \\
& y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left((-\ln 2 \cdot(4 x+3)+4) 2^{-x}\right)^{\prime}= \\
& =\left((-\ln 2 \cdot(4 x+3)+4) 2^{-x}\right)^{\prime}=(-4 \ln 2) 2^{-x}+(-\ln 2 \cdot(4 x+3)+4) 2^{-x} \cdot \ln 2 \cdot(-1)= \\
& =\left(-4 \ln 2+\ln ^{2} 2 \cdot(4 x+3)-4 \ln 2\right) 2^{-x}=\left(\ln ^{2} 2 \cdot(4 x+3)-8 \ln 2\right) 2^{-x} \\
& y^{\prime \prime \prime}=\left(y^{\prime \prime}\right)^{\prime}=\left(\left(\ln ^{2} 2 \cdot(4 x+3)-8 \ln 2\right) 2^{-x}\right)^{\prime}= \\
& =\left(4 \ln ^{2} 2\right) 2^{-x}+\left(\ln ^{2} 2 \cdot(4 x+3)-8 \ln 2\right) 2^{-x} \cdot \ln 2 \cdot(-1)= \\
& =\left(4 \ln ^{2} 2-\ln ^{3} 2 \cdot(4 x+3)+8 \ln ^{2} 2\right) 2^{-x}=\left(-\ln ^{3} 2 \cdot(4 x+3)+12 \ln ^{2} 2\right) 2^{-x} \\
& y^{(I V)}=\left(y^{\prime \prime \prime}\right)^{\prime}=\left(\left(-\ln ^{3} 2 \cdot(4 x+3)+12 \ln ^{2} 2\right) 2^{-x}\right)^{\prime}= \\
& =\left(-4 \ln ^{3} 2\right) 2^{-x}+\left(-\ln ^{3} 2 \cdot(4 x+3)+12 \ln ^{2} 2\right) 2^{-x} \cdot \ln 2 \cdot(-1)= \\
& =\left(-4 \ln ^{3} 2+\ln ^{4} 2 \cdot(4 x+3)-12 \ln ^{3} 2\right) 2^{-x}= \\
& =\left(\ln ^{4} 2 \cdot(4 x+3)-16 \ln ^{3} 2\right) 2^{-x} \\
& y^{(V)}=\left(y^{(I V)}\right)^{\prime}=\left(\left(\ln ^{4} 2 \cdot(4 x+3)-16 \ln ^{3} 2\right) 2^{-x}\right)^{\prime}= \\
& =\left(4 \ln ^{4} 2\right) 2^{-x}+\left(\ln ^{4} 2 \cdot(4 x+3)-16 \ln ^{3} 2\right) 2^{-x} \cdot \ln 2 \cdot(-1)= \\
& =\left(-\ln ^{5} 2 \cdot(4 x+3)+20 \ln ^{4} 2\right) 2^{-x}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 19-12