## Task Condition

Approximately calculate using the differential.

$$
y=\sqrt[3]{x}, x=1,21
$$
## Solution

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

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

Choose:

$x_{0} = 1$

Then:

$\Delta x = 0.21$

## Calculation:

$$
y(1) = \sqrt[3]{1} = 1
$$

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

$y^{\prime}(1) = \frac{1}{3} \cdot 1^{-\frac{2}{3}} = \frac{1}{3}$

We get:

$y(1.21) \approx y(1) + y^{\prime}(1) \cdot 0.21 = 1 + \frac{1}{3} \cdot 0.21 = 1.0^{-}$

## Problem Kuznetsov Differentiation 5-10