Answer
The absolute value of the difference in the two results: $0.0017$
Work Step by Step
$f(x) = x^{\frac{1}{2}}$
$f'(x)=\frac{1}{2}x^{\frac{-1}{2}}$
$f(x + \Delta x) \approx f(x) + f'(x)dx=\sqrt x + \frac{1}{2\sqrt x}dx$
$f(0.99)=f(0.88+0.11) \approx \sqrt 0.88 + \frac{1}{2\sqrt 0.88}(0.11)=\frac{\sqrt 22}{5}+\frac{\sqrt 22}{80}\approx 0.9967$
The true value of this, found using a calculator, is $\sqrt 23 \approx 0.995$
The absolute value of the difference in the two results: $0.0017$