## Task Condition

Approximately calculate using the differential.

$y=\sqrt{4 x-1}, x=2.56$
## 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:

$x_{0} = 2.5$

Then:

$\Delta x = 0.06$

Calculate:
$y(2.5) = \sqrt{4 \cdot 2.5 - 1} = 3$

$y^{\prime} = (\sqrt{4 x - 1})^{\prime} = \frac{1}{2 \sqrt{4 x - 1}} \cdot 4 = \frac{2}{\sqrt{4 x - 1}}$

$y^{\prime}(2.5) = \frac{2}{\sqrt{4 \cdot 2.5 - 1}} = \frac{2}{3}$

We get:

$y(2.56) \approx y(2.5) + y^{\prime}(2.5) \cdot 0.06 = 3 + \frac{2}{3} \cdot 0.06 = 3.04$

## Problem Kuznetsov Differentiation 5-17