SOLUTION 1: Begin with the function
$$f(x)= \sqrt{x}$$
and choose
$$x-values: 25 \rightarrow 28$$
so that
$$\Delta x = 28-25 = 3$$
The derivative of $\ y=f(x) \$ is
$$f'(x)= \displaystyle{ 1 \over 2 }x^{-1/2} = \displaystyle{ 1 \over 2 \sqrt{x} }$$
The exact change of $y-$values is
$$\Delta y = f(28) - f(25)$$ $$= \sqrt{28} - \sqrt{25}$$ $$= \sqrt{28} - 5$$
The Differential is
$$dy = f'(25) \ \Delta x$$ $$= \displaystyle{ 1 \over 2 \sqrt{25} } \cdot (3)$$ $$= \displaystyle{ 1 \over 2 (5) } (3)$$ $$= \displaystyle{ 1 \over 10 } (3)$$ $$= 0.3$$
We will assume that
$$\Delta y \approx dy \ \ \ \ \longrightarrow$$ $$\sqrt{28} - 5 \approx 0.3 \ \ \ \ \longrightarrow$$ $$\sqrt{28} \approx 5 + 0.3 \ \ \ \ \longrightarrow$$ $$\sqrt{28} \approx 5.3$$

NOTE: The number 25 was chosen for its proximity to 28 and for it's convenient square root. Check the accuracy of the final estimate using a CALCULATOR: $\sqrt{28} \approx 5.2915$