SOLUTION 2: Begin with the function
$$ f(x)= x^{1/3} $$
and choose
$$ x-values: 8 \rightarrow 10 $$
so that
$$ \Delta x = 10-8 = 2 $$
The derivative of $ \ y=f(x) \ $ is
$$ f'(x)= \displaystyle{ 1 \over 3 }x^{-2/3} = \displaystyle{ 1 \over 3 x^{2/3} } $$
The exact change of $y-$values is
$$ \Delta y = f(10) - f(8) $$
$$ = 10^{1/3} - 8^{1/3} $$
$$ = 10^{1/3} - 2 $$
The Differential is
$$ dy = f'(8) \ \Delta x $$
$$ = \displaystyle{ 1 \over 3 (8)^{2/3} } \cdot (2) $$
$$ = \displaystyle{ 1 \over 3 ((8)^{1/3})^2 } \cdot (2) $$
$$ = \displaystyle{ 1 \over 3 (2)^2 } \cdot (2) $$
$$ = \displaystyle{ 1 \over 3 (4) } (2) $$
$$ = \displaystyle{ 1 \over 12 } (2) $$
$$ = \displaystyle{1 \over 6} $$
We will assume that
$$ \Delta y \approx dy \ \ \ \ \longrightarrow $$
$$ 10^{1/3} - 2 \approx \displaystyle{1 \over 6} \ \ \ \ \longrightarrow $$
$$ 10^{1/3} \approx 2 + \displaystyle{1 \over 6} \ \ \ \ \longrightarrow $$
$$ 10^{1/3} \approx \displaystyle{ 12 \over 6 } + \displaystyle{1 \over 6} \ \ \ \ \longrightarrow $$
$$ 10^{1/3} \approx \displaystyle{13 \over 6} ( \approx 2.1667 ) $$
NOTE: The number 8 was chosen for its proximity to 10 and for it's convenient cube root. Check the accuracy of the final estimate using a CALCULATOR: $ 10^{1/3} \approx 2.1544 $
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