Step 1:
Let $\Delta y=\sqrt{x+\Delta x}-\sqrt x$
$\Delta y=\sqrt{x+\Delta x}-\sqrt x$
$\quad\;=\sqrt{49+.5}-\sqrt {49}$
$\quad\;=\sqrt{49.5}-\sqrt {49}$
$\quad\;=\sqrt{49.5}-7$
$\sqrt{49.5}=7+\Delta y$------(1)
Step 2:
$dy$ is the approximate value of $\Delta y$
$dy=\big(\large\frac{dy}{dx}\big)$$\Delta x$
We know that $y=\sqrt x$
$\large\frac{dy}{dx}=\frac{1}{2\sqrt x}$[Differentiating with respect to $x$]
$dy=\large\frac{dy}{dx}$$.\Delta x$
$\quad\;=\large\frac{1}{2\sqrt x}$$.\Delta x$
$\quad\;=\large\frac{1}{2\sqrt {49}}$$.0.5$
$\quad\;=\large\frac{1}{2\times 7}$$.0.5$
$\quad\;=\large\frac{1}{14}$$\times 0.5$
$\quad\;=0.036$
Step 3:
$\sqrt{49.5}=7+\Delta y$[From equation (1) we have]
Now substitute the value of $dy$ in (1)
$\sqrt{49.5}=7+0.036$
$\qquad\;\;=7.036$