## Task Condition

Approximately calculate using the differential.

$y=\sqrt[3]{x}, x=7.64$
## 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:

$x_{0} = 8$

Then:

$\Delta x = -0.36$

Calculate:

$y(8) = \sqrt[3]{8} = 2$

$y^{\prime} = (\sqrt[3]{x})^{\prime} = \left(x^{\frac{1}{3}}\right)^{\prime} = \frac{1}{3} \cdot x^{-\frac{2}{3}}$
$y^{\prime}(8) = \frac{1}{3} \cdot 8^{-\frac{2}{3}} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$

We get:

$y(7.64) \approx y(8) + y^{\prime}(8) \cdot (-0.36) = 2 - \frac{1}{12} \cdot 0.36 = 1.97$

## Problem Kuznetsov Differentiation 5-16