SOLUTION 16: Begin with the function
$$ f(x)= (8+x)^{1/3} $$
and choose
$$ x-values: 0 \rightarrow 5h^3 $$
so that
$$ \Delta x = 5h^3 - 0 = 5h^3 $$
The derivative of $ \ y=f(x) \ $ is
$$ f'(x)= \displaystyle{ 1 \over 3 }(8+x)^{-2/3} = { 1 \over 3(8+x)^{2/3} } $$
The exact change of $y-$values is
$$ \Delta y = f(5h^3) - f(0) $$
$$ = (8+(5h^3))^{1/3} - (8+(0))^{1/3} $$
$$ = (8+5h^3)^{1/3} - (8)^{1/3} $$
$$ = (8+5h^3)^{1/3} - 2 $$
The Differential is
$$ dy = f'(0) \ \Delta x $$
$$ = \displaystyle{ 1 \over 3(8+(0))^{2/3} } \cdot (5h^3) $$
$$ = \displaystyle{ 1 \over 3((8)^{1/3})^2 } \cdot (5h^3) $$
$$ = \displaystyle{ 1 \over 3(2)^2 } \cdot (5h^3) $$
$$ = \displaystyle{ 1 \over 3(4) } \cdot (5h^3) $$
$$ = \displaystyle{ 5 \over 12 } h^3 $$
Since $h$ is "small" we will assume that
$$ \Delta y \approx dy \ \ \ \ \longrightarrow $$
$$ (8+5h^3)^{1/3} - 2 \approx \displaystyle{ 5 \over 12 } h^3 $$
$$ (8+5h^3)^{1/3} \approx 2+\displaystyle{ 5 \over 12 } h^3 $$
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