Step 1:
Let $y=\sqrt x$
$x=25$
$\Delta x=0.3$
$\Delta y=\sqrt{x+\Delta x}-\sqrt x$
$\quad\;=\sqrt{25+0.3}-\sqrt{25}$
$\quad\;=\sqrt{25.3}-\sqrt{25}$
$\quad\;=\sqrt{25.3}-5$
$\sqrt{25.3}=\Delta y+5$-------(1)
Now $dy$ is approximately equal to $\Delta y$
$dy=\big(\large\frac{dy}{dx}\big)\Delta x$
$\large\frac{dy}{dx}=\frac{1}{2\sqrt x}$[differentiating y with respect to x]
$dy=\large\frac{1}{2\sqrt x}$$\Delta x$
Step 2:
By substituting the value of $\sqrt x$ and $\Delta x$ we get,
$\Rightarrow \large\frac{1}{2\sqrt {25}}$$\times 0.3$
$\Rightarrow \large\frac{1}{2\times 5}$$\times 0.3$
$\Rightarrow \large\frac{1}{10}$$\times 0.3$
$dy=\large\frac{0.3}{10}$$=0.03$
Substitute the value of $dy$ in equ(1)
$\sqrt{25.3}=5+\Delta y$
$\qquad\;\;=5+0.03$
$\sqrt{25.3}=5.03$