## Task Condition

Find the derivative.

$y=x^{e^{\cos x}}$
## Solution

$y=x^{e^{\cos x}}$

$\ln y=e^{\cos x} \cdot \ln x$

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

$=e^{\cos x} \cdot(\cos x)^{\prime} \cdot \ln x+e^{\cos x} \cdot \frac{1}{x}=e^{\cos x} \cdot\left(-\sin x \cdot \ln x+\frac{1}{x}\right)$
$y^{\prime}=y \cdot e^{\cos x} \cdot\left(\frac{1}{x}-\sin x \cdot \ln x\right)=x^{e^{\cos x}} \cdot e^{\cos x} \cdot\left(\frac{1}{x}-\sin x \cdot \ln x\right)$

## Problem Kuznetsov Differentiation 12-23