## Task Condition

Approximately calculate using the differential.

$y=\sqrt[3]{x^{2}+2 x+5}, x=0.97$
## Solution

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

$$
\begin{aligned}
& y(1) = \sqrt[3]{1^{2} + 2 \cdot 1 + 5} = \sqrt[3]{8} = 2 \\
& y^{\prime} = \left(\sqrt[3]{x^{2} + 2 x + 5}\right)^{\prime} = \left(\left(x^{2} + 2 x + 5\right)^{\frac{1}{3}}\right)^{\prime} = \frac{1}{3} \cdot \left(x^{2} + 2 x + 5\right)^{-\frac{2}{3}} \cdot (2 x + 2) = \\
& = \frac{2(x + 1)}{3 \sqrt[3]{\left(x^{2} + 2 x + 5\right)^{2}}} \\
& y^{\prime}(1) = \frac{2(1 + 1)}{3 \sqrt[3]{\left(1^{2} + 2 \cdot 1 + 5\right)^{2}}} = \frac{4}{3 \sqrt[3]{8^{2}}} = \frac{1}{3}
\end{aligned}
$$

We get:

$y(0.97) \approx y(1) + y^{\prime}(1) \cdot (-0.03) = 2 - \frac{1}{3} \cdot 0.03 = 1.99$

## Problem Kuznetsov Differentiation 5-6