## Problem Statement

Approximately calculate using the differential.

$y=\sqrt{4 x-3}, x=1.78$
## 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:

$x_{0} = 1.75$

Then:

$\Delta x = 0.03$

Calculate:
$y(1.75) = \sqrt{4 \cdot 1.75 - 3} = 2$

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

$y^{\prime}(1.75) = \frac{2}{\sqrt{4 \cdot 1.75 - 3}} = \frac{2}{2} = 1$

We get:

$y(1.78) \approx y(1.75) + y^{\prime}(1.75) \cdot 0.03 = 2 + 1 \cdot 0.03 = 2.03$

## Problem Kuznetsov Differentiation 2-22