511. Calculate $\operatorname{tg} 46^{\circ}$, based on the value of the function $y=\operatorname{tg} x$ at $x=45^{\circ}$ and replacing its increment with a differential.
Solution. Here $x=45^{\circ}, \Delta x=46^{\circ}-45^{\circ}=1^{\circ}$. To use formula (6), we need to express the angles in radians: $x=\pi / 4, \Delta x=$ $=0.0175$.

We calculate the differential of the function at $x=\pi / 4$ and $d x=\Delta x=0.0175$:

$$
d y=\left.\frac{1}{\cos ^{2} x} d x\right|_{\substack{x=\pi / 4 \\ d x=0.0175}}=\frac{1}{\left(\frac{\sqrt{2}}{2}\right)^{2}} \cdot 0.0175=0.0350
$$

Therefore, $\Delta y \approx 0.0350$.

Now we find

$$
\operatorname{tg} 46^{\circ}=y+\left.\Delta y\right|_{\substack{x=\pi / 4 \\ d x=0.0175}} \approx \operatorname{tg} 45^{\circ}+0.0350=1.0350
$$

Calculations using four-digit tables give: $\operatorname{tg} 46^{\circ}=$ $=1.0355$