12.21 At what angle to the $O x$ axis is the tangent to the graph of the function $g(x)=x^{2} \ln x$ at the point $x_{0}=1$ inclined?
12.21 The slope coefficient of the tangent to the graph of the function $g(x)$ is equal to $k=g^{\prime}\left(x_{0}\right)=\operatorname{tg} \alpha$, where $x_{0}$ is the abscissa of the point of tangency, and $\alpha$ is the angle of inclination of the tangent to the $O x$ axis. We find

$$
g^{\prime}(x)=2 x \ln x+x^{2} \cdot \frac{1}{x}=2 x \ln x+x=x(2 \ln x+1)
$$

For $x_{0}=1$, we get $g^{\prime}(1)=1$, hence $\operatorname{tg} \alpha=1$ and, consequently, $\alpha=\frac{\pi}{4}$.

Answer: $\frac{\pi}{4}$.