## Task Condition

Approximately calculate using the differential.

$y=\sqrt[5]{x^{2}}, x=1.03$
## Solution

If the increment $\Delta x = x - x_{0}$ of the 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:

$\Delta x = 0.03$

Calculate:

$y(1) = \sqrt[5]{1^{2}} = 1$

$$
\begin{aligned}
& y^{\prime} = \left(\sqrt[5]{x^{2}}\right)^{\prime} = \left(x^{\frac{2}{5}}\right)^{\prime} = \frac{2}{5} \cdot x^{-\frac{3}{5}} \\
& y^{\prime}(1) = \frac{2}{5} \cdot 1^{-\frac{3}{5}} = \frac{2}{5} = 0.4
\end{aligned}
$$

We get:

$y(1.03) \approx y(1) + y^{\prime}(1) \cdot 0.03 = 1 + 0.4 \cdot 0.03 = 1.012$

## Problem Kuznetsov Differentiation 5-25