SOLUTION 17: Begin with the function
$$ f(x)= \ln(4+x) $$
and choose
$$ x-values: 0 \rightarrow 7h $$
so that
$$ \Delta x = 7h - 0 = 7h $$
The derivative of $ \ y=f(x) \ $ is
$$ f'(x)= \displaystyle{ 1 \over 4+x } $$
The exact change of $y-$values is
$$ \Delta y = f(7h) - f(0) $$
$$ = \ln(4+(7h)) - \ln(4+(0)) $$
$$ = \ln(4+7h) - \ln(4) $$
The Differential is
$$ dy = f'(0) \ \Delta x $$
$$ = \displaystyle{ 1 \over 4+(0) } \cdot (7h) $$
$$ = \displaystyle{ 1 \over 4 } \cdot (7h) $$
$$ = \displaystyle{ 7 \over 4 } h $$
Since $h$ is "small" we will assume that
$$ \Delta y \approx dy \ \ \ \ \longrightarrow $$
$$ \ln(4+7h) - \ln(4) \approx \displaystyle{ 7 \over 4 } h \ \ \ \ \longrightarrow $$
$$ \ln(4+7h) \approx \ln(4) + \displaystyle{ 7 \over 4 } h $$
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