Condition of the problem

Calculate approximately using the differential.

$y=\sqrt[3]{x^{3}+7 x}, x=1,012$
## 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:
$\Delta x = 0.012$

Calculate:

$y(1) = \sqrt[3]{1^{3} + 7 \cdot 1} = \sqrt[3]{8} = 2$

$y^{\prime} = \left(\sqrt[3]{x^{3} + 7 x}\right)^{\prime} = \left(\left(x^{3} + 7 x\right)^{\frac{1}{3}}\right)^{\prime} = \frac{1}{3} \cdot \left(x^{3} + 7 x\right)^{-\frac{2}{3}} \cdot \left(x^{3} + 7 x\right)^{\prime} = $

$= \frac{1}{3} \cdot \left(x^{3} + 7 x\right)^{-\frac{2}{3}} \cdot \left(3 x^{2} + 7\right) = \frac{3 x^{2} + 7}{3 \sqrt[3]{\left(x^{3} + 7 x\right)^{2}}}$

$y^{\prime}(1) = \frac{3 \cdot 1^{2} + 7}{3 \sqrt[3]{\left(1^{3} + 7 \cdot 1\right)^{2}}} = \frac{10}{3 \sqrt[3]{8^{2}}} = \frac{10}{3 \cdot 4} = \frac{5}{6}$

We get:

$y(1.012) \approx y(1) + y^{\prime}(1) \cdot 0.012 = 2 + \frac{5}{6} \cdot 0.012 = 2 + 0.01 = 2.01$

## Problem Kuznetsov Differentiation 5-2