SOLUTION 8: Begin with the function
$$f(x)= x^{2/5}$$
and choose
$$x-values: 32 \rightarrow 30$$
so that
$$\Delta x = 30-32 = -2$$
The derivative of $\ y=f(x) \$ is
$$f'(x)= \displaystyle{ 2 \over 5 }x^{-3/5} = \displaystyle{ 2 \over 5 x^{3/5} }$$
The exact change of $y-$values is
$$\Delta y = f(30) - f(32)$$ $$= 30^{2/5} - 32^{2/5}$$ $$= 30^{2/5} - (32^{1/5})^2$$ $$= 30^{2/5} - (2)^2$$ $$= 30^{2/5} - 4$$
The Differential is
$$dy = f'(32) \ \Delta x$$ $$= \displaystyle{ 2 \over 5 (32)^{3/5} } \cdot (-2)$$ $$= \displaystyle{ 2 \over 5 ((32)^{1/5})^3 } \cdot (-2)$$ $$= \displaystyle{ 1 \over 3 (2)^3 } \cdot (-2)$$ $$= \displaystyle{ 1 \over 3 (8) } (-2)$$ $$= \displaystyle{ 1 \over 24 } (-2)$$ $$= \displaystyle{-1 \over 12}$$
We will assume that
$$\Delta y \approx dy \ \ \ \ \longrightarrow$$ $$30^{2/5} - 4 \approx \displaystyle{ -1 \over 12 } \ \ \ \ \longrightarrow$$ $$30^{2/5} \approx 4 +\displaystyle{ -1 \over 12 } \ \ \ \ \longrightarrow$$ $$30^{2/5} \approx \displaystyle{ 48 \over 12 } + \displaystyle{ -1 \over 12 } \ \ \ \ \longrightarrow$$ $$30^{2/5} \approx \displaystyle{ 47 \over 12} ( \approx 3.9167 )$$

NOTE: The number 32 was chosen for its proximity to 30 and for it's convenient fifth root. Check the accuracy of the final estimate using a CALCULATOR: $30^{2/5} \approx 3.8981$