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: $10^{1/3} \approx 2.1544$