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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