## Task Condition

Find the differential $d y$.

$$
y=\arccos \left(\frac{x^{2}-1}{x^{2} \sqrt{2}}\right)
$$
## Solution

$d y=y^{\prime} \cdot d x=\left(\arccos \left(\frac{x^{2}-1}{x^{2} \sqrt{2}}\right)\right)^{\prime} d x=\frac{-1}{\sqrt{1-\left(\frac{x^{2}-1}{x^{2} \sqrt{2}}\right)^{2}}} \cdot\left(\frac{x^{2}-1}{x^{-2} \sqrt{2}}\right)^{\prime} d x=$
$=\frac{-1}{\sqrt{\frac{2 x^{4}}{2 x^{4}}-\frac{\left(x^{2}-1\right)^{2}}{2 x^{4}}}} \cdot \frac{2 x \cdot x^{2}-\left(x^{2}-1\right) \cdot 2 x}{\sqrt{2} x^{4}} d x=\frac{-\sqrt{2} \cdot x^{2}}{\sqrt{2 x^{4}-x^{4}+2 x^{2}-1}} \cdot \frac{2 x^{3}-2 x^{3}+2 x}{\sqrt{2} x^{4}} d x=$
$=\frac{-\sqrt{2} \cdot x^{2}}{\sqrt{x^{4}+2 x^{2}-1}} \cdot \frac{2 x}{\sqrt{2} x^{4}} d x=-\frac{2}{x \sqrt{x^{4}+2 x^{2}-1}} d x$

## Problem Kuznetsov Differentiation 4-8

## Problem Statement

## u

Approximate using the differential.

$$
y=\sqrt{x^{2}+x+3}, x=1.9
$$

## Solution

If the increment $\Delta x=x-x_{\text {■argument }} x$ is small in absolute value, then

$$
f(x)=f\left(x_{0}+\Delta x\right) \approx f\left(x_{\bar{U}}\right)+f^{\prime}\left(x_{\bar{U}}\right) \cdot \Delta x
$$

Choose:

$x_{0}=2$

Then:

$\Delta x=-0.03$

Calculate:

$$
\begin{aligned}
& y(2)=\sqrt{2^{2}+2+3}=3 \\
& y^{\prime}=\left(\sqrt{x^{2}+x+3}\right)^{\prime}=\left(\frac{1}{2 \sqrt{x^{2}+x+3}} \cdot(2 x+1)\right)^{\prime}=\frac{2 x+1}{2 \sqrt{x^{2}+x+3}} \\
& y^{\prime}(2)=\frac{2 \cdot 2+1}{2 \sqrt{2^{2}+2+3}}=\frac{5}{6}
\end{aligned}
$$

We get:

$$
y(1.97) \approx y(2)+y^{\prime}(2) \cdot(-0.03)=3-\frac{5}{6} \cdot 0.03=2.975
$$

## Problem Kuznetsov Differentiation 5-8