## Problem Statement

Calculate approximately using the differential.

$$
y=\frac{1}{\sqrt{2 x^{2}+x+1}}, x=1.016
$$
## 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:

$\Delta x = 0.016$

Calculate:

\[
\begin{aligned}
& y(1) = \frac{1}{\sqrt{2 \cdot 1^{2} + 1 + 1}} = \frac{1}{2} \\
& y^{\prime} = \left(\frac{1}{\sqrt{2 x^{2} + x + 1}}\right)^{\prime} = \left((2 x^{2} + x + 1)^{-\frac{1}{2}}\right)^{\prime} = -\frac{1}{2} \cdot (2 x^{2} + x + 1)^{-\frac{3}{2}} \cdot (4 x + 1) \\
& y^{\prime}(1) = -\frac{1}{2} \cdot (2 \cdot 1^{2} + 1 + 1)^{-\frac{3}{2}} \cdot (4 \cdot 1 + 1) = -\frac{1}{2} \cdot \frac{1}{8} \cdot 5 = -\frac{5}{16}
\end{aligned}
\]

We get:

$y(1.016) \approx y(1) + y^{\prime}(1) \cdot 0.016 = \frac{1}{2} - \frac{5}{16} \cdot 0.016 = 0.495$

## Problem Kuznetsov Differentiation 5-18