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
$$\Delta y \approx dy \ \ \ \ \longrightarrow$$ $$\displaystyle{ 8-h^2 \over (1+h^2)^2 } - 8 \approx -17h^2$$ $$\displaystyle{ 8-h^2 \over (1+h^2)^2 } \approx 8-17h^2$$