## Task Condition

Approximately calculate using the differential.

$y=\sqrt[4]{2 x-\sin \frac{\pi x}{2}}, x=1.02$
## 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:

$\Delta x=0.02$

Calculate:

$$
\begin{aligned}
& y(1)=\sqrt[4]{2 \cdot 1-\sin \frac{\pi \cdot 1}{2}}=\sqrt[4]{2-1}=1 \\
& y^{\prime}=\left(\sqrt[4]{2 x-\sin \frac{\pi x}{2}}\right)^{\prime}=\left(\left(2 x-\sin \frac{\pi x}{2}\right)^{\frac{1}{4}}\right)^{\prime}=\frac{1}{4} \cdot\left(2 x-\sin \frac{\pi x}{2}\right)^{-\frac{3}{4}} \cdot\left(2-\cos \frac{\pi x}{2} \cdot \frac{\pi}{2}\right) \\
& y^{\prime}(1)=\frac{1}{4} \cdot\left(2 \cdot 1-\sin \frac{\pi \cdot 1}{2}\right)^{-\frac{3}{4}} \cdot\left(2-\cos \frac{\pi \cdot 1}{2} \cdot \frac{\pi}{2}\right)=\frac{1}{4} \cdot 1^{-\frac{3}{4}} \cdot\left(2-0 \cdot \frac{\pi}{2}\right)=\frac{1}{2}
\end{aligned}
$$

We get:

$$
y(1.02) \approx y(1)+y^{\prime}(1) \cdot 0.02=1+\frac{1}{2} \cdot 0.02=1.01
$$