SOLUTION 20: Begin with the function
$$ f(x)= \displaystyle{ 1 \over \sqrt{1-x} } $$
and choose
$$ x-values: 0 \rightarrow h^2 $$
so that
$$ \Delta x = h^2 - 0 = h^2 $$
The derivative of $ \ y=f(x) \ $ is
$$ f'(x)= D \Big\{ \displaystyle{ 1 \over \sqrt{1-x} } \Big\} = D\{(1-x)^{-1/2} \}
= { -1 \over 2 } (1-x)^{-3/2} \cdot (-1) = { 1 \over 2(1-x)^{3/2} } $$
The exact change of $y-$values is
$$ \Delta y = f(h^2 ) - f(0) $$
$$ = \displaystyle{ 1 \over \sqrt{1-(h^2)} } - \displaystyle{ 1 \over \sqrt{1-(0)} } $$
$$ = \displaystyle{ 1 \over \sqrt{1-h^2} } - \displaystyle{ 1 \over \sqrt{1} } $$
$$ = \displaystyle{ 1 \over \sqrt{1-h^2} } - 1 $$
The Differential is
$$ dy = f'(0) \ \Delta x $$
$$ = \displaystyle { 1 \over 2(1-(0))^{3/2} } \cdot (h^2) $$
$$ = \displaystyle { 1 \over 2(1)^{3/2} } \cdot (h^2) $$
$$ = \displaystyle { 1 \over 2(1) } \cdot (h^2) $$
$$ = \displaystyle{ 1 \over 2 }h^2 $$
Since $h$ is "small" we will assume that
$$ \Delta y \approx dy \ \ \ \ \longrightarrow $$
$$ \displaystyle{ 1 \over \sqrt{1-h^2} } - 1 \approx \displaystyle{ 1 \over 2 }h^2 \ \ \ \ \longrightarrow $$
$$ \displaystyle{ 1 \over \sqrt{1-h^2} } \approx 1+ \displaystyle{ 1 \over 2 }h^2 \ \ \ \ \longrightarrow $$
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