Wiskunde & ICT - documentatiecentrum

\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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