## Task Condition

Calculate approximately using the differential.

$y=\frac{x+\sqrt{5-x^{2}}}{2}, x=0.98$
## 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:

$x_{0} = 1$

Then:

$\Delta x = -0.02$

Calculate:

\[
\begin{aligned}
& y(1) = \frac{1 + \sqrt{5 - 1^{2}}}{2} = \frac{1 + 2}{2} = \frac{3}{2} \\
& y^{\prime} = \left(\frac{x + \sqrt{5 - x^{2}}}{2}\right)^{\prime} = \frac{1 + \frac{1}{2 \sqrt{5 - x^{2}}} \cdot (-2 x)}{2} = \frac{\sqrt{5 - x^{2}} - x}{2 \sqrt{5 - x^{2}}} \\
& y^{\prime}(1) = \frac{\sqrt{5 - 1^{2}} - 1}{2 \sqrt{5 - 1^{2}}} = \frac{2 - 1}{2 \cdot 2} = \frac{1}{4}
\end{aligned}
\]

We get:
$y(0.98) \approx y(1) + y^{\prime}(1) \cdot (-0.02) = \frac{3}{2} - \frac{1}{4} \cdot 0.02 = 1.5 - 0.005 = 1.495$

## Problem Kuznetsov Differentiation 5-3