## problem statement

Approximately calculate using the differential.

$y=\sqrt{x^{2}+5}, x=1.97$
## 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:

$x_{0}=2$

Then:

$\Delta x=-0.03$

Calculate:

$y(2)=\sqrt{2^{2}+5}=\sqrt{9}=3$

$y^{\prime}=\left(\sqrt{x^{2}+5}\right)^{\prime}=\frac{1}{2 \sqrt{x^{2}+5}} \cdot 2 x=\frac{x}{\sqrt{x^{2}+5}}$

$y^{\prime}(2)=\frac{2}{\sqrt{2^{2}+5}}=\frac{2}{3}$

We get:

$y(1.97) \approx y(2)+y^{\prime}(2) \cdot(-0.03)=3-\frac{2}{3} \cdot 0.03=2.98$