## Task Condition

Calculate approximately using the differential.

$$
y=\frac{1}{\sqrt{x}}, x=4,16
$$
## 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:

$x_{0} = 4$

Then:

$\Delta x = 0.16$

Calculate:

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

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

$y^{\prime}(4) = -\frac{1}{2} \cdot 4^{-\frac{3}{2}} = -\frac{1}{2} \cdot \frac{1}{8} = -\frac{1}{16}$

We get:

$y(4.16) \approx y(4) + y^{\prime}(4) \cdot 0.16 = \frac{1}{2} - \frac{1}{16} \cdot 0.16 = 0.49$

## Problem Kuznetsov Differentiation 5-20