SOLUTION 18: Begin with the function
$$ f(x)= \log(100-x) $$
and choose
$$ x-values: 0 \rightarrow h^4 $$
so that
$$ \Delta x = h^4 - 0 = h^4 $$
$\Big($Recall that $ D\{ \log(g(x)) \} = \displaystyle{ 1 \over g(x) } \cdot g'(x) \cdot { 1 \over \ln(10) }. \Big) $
The derivative of $ \ y=f(x) \ $ is
$$ f'(x)= \displaystyle{ 1 \over 100-x } \cdot (-1) \cdot { 1 \over \ln(10) } $$
$$ = \displaystyle{ -1 \over (-1)(x-100) } \cdot { 1 \over \ln(10) } $$
$$ = \displaystyle{ 1 \over x-100 } \cdot { 1 \over \ln(10) } $$
The exact change of $y-$values is
$$ \Delta y = f(h^4) - f(0) $$
$$ = \log(100-(h^4)) - \log(100+(0)) $$
$$ = \log(100-h^4) - \log(100) $$
$\Big($ Recall that $ \log(10^n)=n .\Big)$
$$ = \log(100-h^4) - \log(10^2) $$
$$ = \log(100-h^4) - 2 $$
The Differential is
$$ dy = f'(0) \ \Delta x $$
$$ = \displaystyle{ 1 \over ((0)^4) - 100 } \cdot { 1 \over \ln(10) } \cdot (h^4) $$
$$ = \displaystyle{ 1 \over -100 } \cdot { 1 \over \ln(10) } \cdot (h^4) $$
$$ = \displaystyle{ -1 \over 100 \ln(10) } h^4 $$
Since $h$ is "small" we will assume that
$$ \Delta y \approx dy \ \ \ \ \longrightarrow $$
$$ \log(100-h^4) - 2 \approx \displaystyle{ -1 \over 100 \ln(10) } h^4 \ \ \ \ \longrightarrow $$
$$ \log(100-h^4) \approx 2 - \displaystyle{ 1 \over 100 \ln(10) } h^4 $$
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