## Task Condition

Find the derivative.

$y=x^{29^{x}} \cdot 29^{x}$
## Solution

$y=x^{29^{x}} \cdot 29^{x}$

$\ln y=\ln \left(x^{29^{x}} \cdot 29^{x}\right)=\ln \left(x^{29^{x}}\right)+x \ln 29=29^{x} \cdot \ln (x)+x \ln 29$
$\frac{y^{\prime}}{y}=\left(29^{x} \cdot \ln (x)+x \ln 29\right)^{\prime}=29^{x} \cdot \ln 29 \cdot \ln (x)+29^{x} \cdot \frac{1}{x}+\ln 29$

$y^{\prime}=y \cdot\left(29^{x} \cdot \ln 29 \cdot \ln (x)+29^{x} \cdot \frac{1}{x}+\ln 29\right)=x^{29^{x}} \cdot 29^{x} \cdot\left(29^{x} \cdot \ln 29 \cdot \ln (x)+\frac{29^{x}}{x}+\ln 29\right)$

## Problem Kuznetsov Differentiation 12-29