## Task Condition

Approximately calculate using the differential.

$y=\sqrt[3]{3 x+\cos x}, x=0.01$
## 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:

$x_{0} = 0$

Then:

$\Delta x = 0.01$

Calculate:

$y(0) = \sqrt[3]{3 \cdot 0 + \cos 0} = \sqrt[3]{0 + 1} = 1$

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

$y^{\prime}(0) = \frac{1}{3} \cdot (3 \cdot 0 + \cos 0)^{-\frac{2}{3}} \cdot (3 - \sin 0) = \frac{1}{3} \cdot 1^{-\frac{2}{3}} \cdot 3 = 1$

We get:

$y(0.01) \approx y(1) + y^{\prime}(1) \cdot 0.01 = 1 + 1 \cdot 0.01 = 1.01$

## Problem Kuznetsov Differentiation 5-28