## Task Condition

Approximately calculate using the differential.

$y=\sqrt{x^{3}}, x=0.98$
## 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:

$\Delta x = -0.02$

Calculate:

$y(1) = \sqrt{1^{3}} = 1$

$y^{\prime} = \left(\sqrt{x^{3}}\right)^{\prime} = \frac{1}{2 \sqrt{x^{3}}} \cdot 3 x^{2} = \frac{3 \sqrt{x}}{2}$

$y^{\prime}(1) = \frac{3 \sqrt{1}}{2} = \frac{3}{2}$

We get:

$y(0.98) \approx y(1) + y^{\prime}(1) \cdot (-0.02) = 1 - \frac{3}{2} \cdot 0.02 = 0.97$

## Problem Kuznetsov Differentiation 2-23