Answer
$$\sqrt {1 + \Delta x} \approx 1 + \frac{1}{2}\Delta x$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sqrt x ;{\text{ }}\sqrt {1 + \Delta x} \approx 1 + \frac{1}{2}\Delta x,{\text{ }}{x_0} = 1,{\text{ where }}\Delta x = x - 1 \cr
& {\text{Calculate }}f'\left( x \right) \cr
& f'\left( x \right) = \frac{1}{{2\sqrt x }} \cr
& {\text{Evaluate }}f\left( {{x_0}} \right){\text{ and }}f'\left( {{x_0}} \right){\text{ }} \cr
& f\left( {{x_0}} \right) = \sqrt 1 = 1 \cr
& f'\left( {{x_0}} \right) = \frac{1}{{2\sqrt 1 }} = \frac{1}{2} \cr
& {\text{Apply }}f\left( {{x_0} + \Delta x} \right) \approx f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\Delta x{\text{ and substitute the known values}} \cr
& {\text{ }}\sqrt {1 + \Delta x} \approx 1 + \frac{1}{2}\Delta x \cr} $$
