501. Using the differential, calculate the increment of the function $y=x \sqrt{x^{2}+5}$ with an accuracy of 0.01 when $x=2$ and $\Delta x=0.2$.
Solution. We find the differential of the given function:

$$
d y=y^{\prime} d x=\left(\sqrt{x^{2}+5}+\frac{x^{2}}{\sqrt{x^{2}+5}}\right) d x
$$

For $x=2$ and $\Delta x=0.2$, we get

$$
\left.\Delta y \approx d y\right|_{\substack{x=2 . \\ d x=0,2}}=\left(\sqrt{4+5}+\frac{4}{\sqrt{4+5}}\right) \cdot 0.2 \approx 0.866 \approx 0.87
$$