## Task Condition

Approximately calculate using the differential.

$y=\frac{1}{\sqrt{2 x+1}}, x=1.58$
## Solution

If the increment $\Delta x = x - x_{0 \text{ of the argument }} x$ 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:

$x_{0} = 1.5$

Then:

$\Delta x = 0.08$

Calculate:

$y(1.5) = \frac{1}{\sqrt{2 \cdot 1.5 + 1}} = \frac{1}{\sqrt{4}} = \frac{1}{2}$

\[
\begin{aligned}
& y^{\prime} = \left(\frac{1}{\sqrt{2 x + 1}}\right)^{\prime} = \left((2 x + 1)^{-\frac{1}{2}}\right)^{\prime} = -\frac{1}{2} \cdot (2 x + 1)^{-\frac{3}{2}} \cdot 2 = \\
& = -\frac{1}{\sqrt{(2 x + 1)^{3}}} \\
& y^{\prime}(1.5) = -\frac{1}{\sqrt{(2 \cdot 1.5 + 1)^{3}}} = -\frac{1}{\sqrt{4^{3}}} = -\frac{1}{8}
\end{aligned}
\]

We get:

$y(1.58) \approx y(1.5) + y^{\prime}(1.5) \cdot 0.08 = \frac{1}{2} - \frac{1}{8} \cdot 0.08 = 0.5 - 0.01 = 0.49$

## Problem Kuznetsov Differentiation 5-31