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
$$\Delta y \approx dy \ \ \ \ \longrightarrow$$ $$\displaystyle{ h^2 \over 4 + h^2 } \approx \displaystyle{ 1 \over 4 } h^2$$