Antwoorden

$
\eqalign{
  & I.  \cr
  & f(x) = 2{}^{10}\log \left( {1 - x} \right)  \cr
  & f'(x) = 2\frac{1}
{{\left( {1 - x} \right) \cdot \ln (10)}} \cdot  - 1 = \frac{{ - 2}}
{{\left( {1 - x} \right) \cdot \ln (10)}} = \frac{2}
{{\left( {x - 1} \right) \cdot \ln (10)}}  \cr
  & II.  \cr
  & f(x) = \frac{{{}^{\frac{1}
{2}}\log x}}
{{1 - {}^{\frac{1}
{2}}\log x}} = \frac{{\frac{{\ln (x)}}
{{\ln \left( {\frac{1}
{2}} \right)}}}}
{{1 - \frac{{\ln (x)}}
{{\ln \left( {\frac{1}
{2}} \right)}}}} = \frac{{\ln (x)}}
{{\ln \left( {\frac{1}
{2}} \right) - \ln (x)}} = \frac{{\ln (x)}}
{{ - \ln \left( 2 \right) - \ln (x)}} = \frac{{ - \ln (x)}}
{{\ln (2x)}}  \cr
  & f'(x) = \frac{{ - \frac{1}
{x} \cdot \ln (2x) - \left( { - \ln (x) \cdot \frac{1}
{{2x}} \cdot 2} \right)}}
{{\left( {\ln (2x} \right)^2 }} = \frac{{ - \frac{{\ln (2x)}}
{x} + \frac{{\ln (x)}}
{x}}}
{{\left( {\ln (2x} \right)^2 }} = \frac{{ - \ln (2x) + \ln (x)}}
{{x \cdot \left( {\ln (2x} \right)^2 }} =  - \frac{{\ln (2)}}
{{x \cdot \left( {\ln (2x} \right)^2 }} \cr}
$
 

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