## Task Condition

Find the derivative.

$y=x^{3^{x}} \cdot 2^{x}$
## Solution

$y=x^{3^{x}} \cdot 2^{x}$

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

$\frac{y^{\prime}}{y}=\left(3^{x} \cdot \ln (x)+x \ln 2\right)^{\prime}=3^{x} \cdot \ln 3 \cdot \ln (x)+3^{x} \cdot \frac{1}{x}+\ln 2$

$y^{\prime}=y \cdot\left(3^{x} \cdot \ln 3 \cdot \ln (x)+3^{x} \cdot \frac{1}{x}+\ln 2\right)=x^{3^{x}} \cdot 2^{x} \cdot\left(3^{x} \cdot \ln 3 \cdot \ln (x)+\frac{3^{x}}{x}+\ln 2\right)$

## Problem Kuznetsov Differentiation 12-20