## Task Condition

Approximately calculate using the differential.
$y=\sqrt[3]{x^{2}}, x=1.03$
## Solution

If the increment $\Delta x = x - x_{\text{argument}}$ is small in absolute value, then

$$
f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{\overline{\mathrm{U}}}\right) + f^{\prime}\left(x_{\overline{\mathrm{U}}}\right) \cdot \Delta x
$$

Choose:

$x_{0} = 1$

Then

$\Delta x = 0.03$

Calculate:

$$
\begin{aligned}
& y(1) = \sqrt[3]{1^{2}} = 1 \\
& y^{\prime} = \left(\sqrt[3]{x^{-2}}\right)^{\prime} = \left(x^{\frac{2}{3}}\right)^{\prime} = \frac{2}{3} \cdot x^{-\frac{1}{3}} \\
& y^{\prime}(1) = \frac{2}{3} \cdot 1^{-\frac{1}{3}} = \frac{2}{3}
\end{aligned}
$$

We get:

$y(1.03) \approx y(1) + y^{\prime}(1) \cdot 0.03 = 1 + \frac{2}{3} \cdot 0.03 = 1 + 0.02 = 1.02$

## Problem Kuznetsov Differentiation 2-12