SOLUTION 21: Begin with the function
$$ f(x)= \displaystyle{ 8-x \over (1+x)^2 } $$
and choose
$$ x-values: 0 \rightarrow h^2 $$
so that
$$ \Delta x = h^2 - 0 = h^2 $$
The derivative of $ \ y=f(x) \ $ is
$$ f'(x)= \displaystyle{ (1+x)^2 \cdot (-1)-(8-x) \cdot 2(1+x)(1) \over (1+x)^4 }
= { (1+x) [-(1+x)-2(8-x)] \over (1+x)^4 }
= { -1-x-16+2x \over (1+x)^3 } = { x-17 \over (1+x)^3 } $$
The exact change of $y-$values is
$$ \Delta y = f(h^2) - f(0) $$
$$ = \displaystyle{ 8-(h^2) \over (1+(h^2))^2 } - \displaystyle{ 8-(0) \over (1+(0))^2 } $$
$$ = \displaystyle{ 8-h^2 \over (1+h^2)^2 } - \displaystyle{ 8 \over (1)^2 } $$
$$ = \displaystyle{ 8-h^2 \over (1+h^2)^2 } - \displaystyle{ 8 \over 1 } $$
$$ = \displaystyle{ 8-h^2 \over (1+h^2)^2 } - 8 $$
The Differential is
$$ dy = f'(0) \ \Delta x $$
$$ = \displaystyle{ (0)-17 \over (1+(0))^3 } \cdot (h^2) $$
$$ = \displaystyle{ -17 \over (1)^3 } \cdot (h^2) $$
$$ = \displaystyle{ -17 \over 1 } \cdot (h^2) $$
$$ = -17h^2 $$
Since $h$ is "small" we will assume that
$$ \Delta y \approx dy \ \ \ \ \longrightarrow $$
$$ \displaystyle{ 8-h^2 \over (1+h^2)^2 } - 8 \approx -17h^2 $$
$$ \displaystyle{ 8-h^2 \over (1+h^2)^2 } \approx 8-17h^2 $$
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