463. Find the increment and differential of the function $y=x^{2}$ at the point $x=2$ for $\Delta x=0.1$.
Solution. $\quad \Delta y=(x+\Delta x)^{2}-x^{2}=x^{2}+2 x \Delta x+(\Delta x)^{2}-x^{2}=2 x \Delta x+(\Delta x)^{2}$;

$$
\begin{aligned}
& \left.\Delta y\right|_{\Delta x=0.1}=2 \cdot 2 \cdot 0.1+(0.1)^{2}=0.4+0.01=0.41 \\
& d y=y^{\prime} \cdot \Delta x=\left(x^{2}\right)^{\prime} \cdot \Delta x=2 x \cdot \Delta x ;\left.\quad d y\right|_{\substack{x=2 \\
\Delta x=0.1}}=2 \cdot 2 \cdot 0.1=0.4
\end{aligned}
$$