Using differentials the approximate value of √25.2, is

Consider the function $y=f(x)=\sqrt{x}$.

Let $x=25$ and $x+\Delta x=25.2$. Then, $\Delta x=25.2-25=0.2$

For $x=25$, we have

$y=\sqrt{25}=5$         [Putting x = 25 in y = $\sqrt{x}$]

Let $d x=\Delta x=0.2$

Now,

$y =\sqrt{x} \Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{x}} \Rightarrow\left(\frac{d y}{d x}\right)_{x=25}=\frac{1}{2(5)}=\frac{1}{10}$

∴  $d y =\frac{d y}{d x} d x$

$\Rightarrow d y=\frac{1}{10}(0.2)=0.02 \Rightarrow \Delta y=0.02$                 $[∵ \Delta y \cong d y]$

Hence, $\sqrt{25.2}=y+\Delta y=5+0.02=5.02$