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 $
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