510. Find the approximate value of the function $y=\sqrt{3 x^{2}+1}$ at $x=1.02$.
Solution. We will use formula (6), i.e., $f(x+\Delta x) \approx$
$\approx f(x)+d y$. In this case, we should take $x=1$. Then $\Delta x=$ $=1.02-1=0.02$. The value of the function at $x=1$ is easily determined: $f(1)=\sqrt{3 \cdot 1^{2}+1}=2$. Next, we find

$$
d y=\left(\sqrt{3 x^{2}+1}\right)^{\prime} d x=\frac{6 x d x}{2 \sqrt{3 x^{2}+1}}=\frac{3 x d x}{\sqrt{3 x^{2}+1}}
$$

from which

$$
\left.d y\right|_{\substack{x=1 \\ \Delta x=0.02}}=\frac{3 \cdot 1 \cdot 0.02}{\sqrt{3 \cdot 1^{2}+1}}=\frac{0.06}{2}=0.03
$$

Therefore,

$$
f(1.02) \approx f(1)+0.03=2+0.03=2.03
$$

If the value of the function at $x=1.02$ is calculated directly, we get $f(1.02)=\sqrt{3 \cdot 1.02^{2}+1}=2.03007 \ldots$, i.e., the difference between the exact value of the function at $x=1.02$ and its approximate value is a very small number: $2.03007-2.03=$ $=0.00007$.

Remarks. 1. To correctly apply the formula for approximate calculation of the function value, one needs to learn how to choose the function $f(x)$. For example, to find $2.08^{6}$, we should take $f(x)=x^{6}, x=2, \Delta x=0.08$; to calculate $\sqrt[3]{9}$, we should take $f(x)=\sqrt[3]{x}, x=8, \Delta x=1$.