## Task Condition

Find the $n$-th order derivative.

$y=2^{3 x+5}$
## Solution

$y=2^{3 x+5}=2^{3 x} \cdot 32=32 \cdot\left(e^{\ln 2}\right)^{3 x}=32 e^{3 x \cdot \ln 2}$

$y^{\prime}=\left(32 e^{3 x \cdot \ln 2}\right)^{\prime}=32 e^{3 x \cdot \ln 2} \cdot 3 \cdot \ln 2$

$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(32 e^{3 x \cdot \ln 2} \cdot 3 \cdot \ln 2\right)^{\prime}=32 e^{3 x \cdot \ln 2} \cdot 3^{2} \cdot \ln ^{2} 2$

Obviously,

$y^{(n)}=32 e^{3 x \cdot \ln 2} \cdot 3^{n} \cdot \ln ^{n} 2=2^{3 x+5} \cdot 3^{n} \cdot \ln ^{n} 2$

## Problem Kuznetsov Differentiation 18-11