SOLUTION 14: Begin with the function
$$ f(x)= \sqrt{16 + x} $$
and choose
$$ x-values: 0 \rightarrow 3h $$
so that
$$ \Delta x = 3h - 0 = 3h $$
The derivative of $ \ y=f(x) \ $ is
$$ f'(x)= \displaystyle{ 1 \over 2 } (16+x)^{-1/2} = { 1 \over 2 \sqrt{16+x} }$$
The exact change of $y-$values is
$$ \Delta y = f(3h) - f(0) $$
$$ = \sqrt{ 16+3h } - \sqrt{16+(0)} $$
$$ = \sqrt{ 16+3h } - 4 $$
The Differential is
$$ dy = f'(0) \ \Delta x $$
$$ = \displaystyle{ 1 \over 2 \sqrt{16+(0)} }\cdot (3h) $$
$$ = \displaystyle{ 1 \over 2 (4) } \cdot (3h) $$
$$ = \displaystyle{ 3 \over 8 } h $$
Since $h$ is "small" we will assume that
$$ \Delta y \approx dy \ \ \ \ \longrightarrow $$
$$ \sqrt{ 16+3h } - 4 \approx \displaystyle{ 3 \over 8 } h \ \ \ \ \longrightarrow $$
$$ \sqrt{ 16+3h } \approx 4 + \displaystyle{ 3 \over 8 } h \ \ \ \ \longrightarrow $$
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