6.19. $f(x)=x^{x}$.
Solution.

$$
\begin{aligned}
& \ln f(x)=\ln x^{x}, \ln f(x)=x \ln x \\
& (\ln f(x))^{\prime}=(x \ln x)^{\prime} \\
& \frac{1}{f(x)} \cdot(f(x))^{\prime}=(x)^{\prime} \cdot \ln x+x \cdot(\ln x)^{\prime} \\
& \frac{1}{f(x)} \cdot(f(x))^{\prime}=1 \cdot \ln x+x \cdot \frac{1}{x}=\ln x+1 \\
& (f(x))^{\prime}=f(x) \cdot(\ln x+1)=x^{x} \cdot(\ln x+1)
\end{aligned}
$$

## CALCULATION OF THE DERIVATIVE OF A FUNCTION GIVEN IMPLICITLY