SOLUTION 3: Begin with the function
$$ f(x)= \sqrt{x} $$
and choose
$$ x-values: 100 \rightarrow 96 $$
so that
$$ \Delta x = 96-100 = -4 $$
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
$$ \Delta y = f(96) - f(100) $$ $$ = \sqrt{96} - \sqrt{100} $$ $$ = \sqrt{96} - 10 $$
The Differential is
$$ dy = f'(100) \ \Delta x $$ $$ = \displaystyle{ 1 \over 2 \sqrt{100} } \cdot (-4) $$ $$ = \displaystyle{ 1 \over 2 (10) } (-4) $$ $$ = \displaystyle{ 1 \over 20 } (-4) $$ $$ = \displaystyle{ -1 \over 5 } $$ $$ = -0.2 $$
We will assume that
$$ \Delta y \approx dy \ \ \ \ \longrightarrow $$ $$ \sqrt{96} - 10 \approx -0.2 \ \ \ \ \longrightarrow $$ $$ \sqrt{96} \approx 10-0.2 $$ $$ \sqrt{96} \approx 9.8 $$

NOTE: The number 100 was chosen for its proximity to 96 and for it's convenient square root. Check the accuracy of the final estimate using a CALCULATOR: $ \sqrt{96} \approx 9.7979 $

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