## Task Condition

Calculate approximately using the differential.

$$
y=\sqrt[3]{x}, x=26.46
$$
## Solution

If the increment $\Delta x = x - x_{\text{argument}} x$ is small in absolute value, then $f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x^{x}$

Choose:

$x_{\bar{u}} = 2 \bar{i}$

Then:

$\Delta x = -0.54$

Calculate:

$$
\begin{aligned}
& y(27) = \sqrt[3]{27} = 3 \\
& y^{\prime} = (\sqrt[3]{x})^{\prime} = \left(x^{\frac{1}{3}}\right)^{\prime} = \frac{1}{3} x^{-\frac{2}{3}} \\
& y^{\prime}(2 \overline{7}) = \frac{1}{3} \cdot 27^{-\frac{2}{3}} = \frac{1}{27}
\end{aligned}
$$

We get:

$y(26.16) \approx y\left(2^{-}\right) + y^{\prime}\left(2^{-}\right) \cdot (-0.54) = 3 - \frac{1}{2^{-}} \cdot 0.54 = 2.98$

## Problem Kuznetsov Differentiation 5-7