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
$$\Delta y \approx dy \ \ \ \ \longrightarrow$$ $$\displaystyle{ 1 \over \sqrt{1-h^2} } - 1 \approx \displaystyle{ 1 \over 2 }h^2 \ \ \ \ \longrightarrow$$ $$\displaystyle{ 1 \over \sqrt{1-h^2} } \approx 1+ \displaystyle{ 1 \over 2 }h^2 \ \ \ \ \longrightarrow$$