SOLUTION 15: Begin with the function
$$ f(x)= \displaystyle{ x \over 4 + x } $$
and choose
$$ x-values: 0 \rightarrow h^2 $$
so that
$$ \Delta x = h^2 - 0 = h^2 $$
The derivative of $ \ y=f(x) \ $ is
$$ f'(x)= \displaystyle{ (4+x)(1)-(x)(1) \over (4+x)^2 } = { 4 \over (4+x)^2 } $$
The exact change of $y-$values is
$$ \Delta y = f(h^2) - f(0) $$
$$ = \displaystyle{ (h^2) \over 4 + (h^2) } - \displaystyle{ (0) \over 4 + (0) } $$
$$ = \displaystyle{ h^2 \over 4 + h^2 } - 0 $$
$$ = \displaystyle{ h^2 \over 4 + h^2 } $$
The Differential is
$$ dy = f'(0) \ \Delta x $$
$$ = \displaystyle{ 4 \over (4+(0))^2 } \cdot (h^2) $$
$$ = \displaystyle{ 4 \over 16 } \cdot (h^2) $$
$$ = \displaystyle{ 1 \over 4 } h^2 $$
Since $h$ is "small" we will assume that
$$ \Delta y \approx dy \ \ \ \ \longrightarrow $$
$$ \displaystyle{ h^2 \over 4 + h^2 } \approx \displaystyle{ 1 \over 4 } h^2 $$
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