513. Find the approximate value of the function $y=\sqrt[3]{x^{2}+2 x}$ at $x=2.004$, based on its exact value at $x_{0}=2$ and replacing $\Delta y$ with $d y$.
Solution. Here the initial value $x_{0}=2$, the incremented value $x=x_{0}+\Delta x=2.004, \Delta x=2.004-2=0.004$.

Let's find the initial value of the function: $\left.y\right|_{x=2}=\sqrt[3]{2^{2}+4}=\sqrt[3]{8}=2$.

Let's find the differential:

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

At $x=2$ and $d x=0.004$ we get $d y=\frac{1}{3} \cdot \frac{1}{4} \cdot 6 \cdot 0.004=0.002$.

Let's calculate the approximate value of the function: $\left.y\right|_{x=2.004}=2+0.002=$ $=2.002$.