SOLUTION 19: Begin with the function
$$f(x)= \sqrt{25 + x}$$
and choose
$$x-values: 0 \rightarrow h^3-h^2$$
so that
$$\Delta x = (h^3-h^2) - 0 = h^3-h^2$$
The derivative of $\ y=f(x) \$ is
$$f'(x)= \displaystyle{ 1 \over 2 } (25+x)^{-1/2} = { 1 \over 2 \sqrt{25+x} }$$
The exact change of $y-$values is
$$\Delta y = f(h^3-h^2 ) - f(0)$$ $$= \sqrt{ 25+(h^3-h^2) } - \sqrt{25+(0)}$$ $$= \sqrt{ 25+h^3-h^2 } - \sqrt{25}$$ $$= \sqrt{ 25+h^3-h^2 } - 5$$
The Differential is
$$dy = f'(0) \ \Delta x$$ $$= \displaystyle{ 1 \over 2 \sqrt{25+(0)} }\cdot (h^3-h^2)$$ $$= \displaystyle{ 1 \over 2 (5) } \cdot (h^3-h^2)$$ $$= \displaystyle{ 1 \over 10 }(h^3-h^2)$$ $$= \displaystyle{ 1 \over 10 }h^3 - \displaystyle{ 1 \over 10 }h^2$$
Since $h$ is "small" we will assume that
$$\Delta y \approx dy \ \ \ \ \longrightarrow$$ $$\sqrt{ 25+h^3-h^2 } - 5 \approx \displaystyle{ 1 \over 10 }h^3 - \displaystyle{ 1 \over 10 }h^2 \ \ \ \ \longrightarrow$$ $$\sqrt{ 25+h^3-h^2 } \approx 5 + \displaystyle{ 1 \over 10 }h^3 - \displaystyle{ 1 \over 10 }h^2 \ \ \ \ \longrightarrow$$

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