Answer
$$f'\left( x \right) = 430\left( {{{0.62}^x}} \right)\left( {\frac{{\left( {6.42 + 3.3\left( {{{1.46}^x}} \right)} \right)\left( {\ln 0.62} \right) - \left( {3.3\left( {{{1.46}^x}} \right)\ln 1.46} \right)}}{{{{\left( {6.42 + 3.3\left( {{{1.46}^x}} \right)} \right)}^2}}}} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{430\left( {{{0.62}^x}} \right)}}{{6.42 + 3.3\left( {{{1.46}^x}} \right)}} \cr
& {\text{calculate the derivative of the function}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {\frac{{430\left( {{{0.62}^x}} \right)}}{{6.42 + 3.3\left( {{{1.46}^x}} \right)}}} \right) \cr
& {\text{Use the quotient rule }}\left( {{\text{see the page 237}}} \right) \cr
& f'\left( x \right) = 430\left( {\frac{{\left( {6.42 + 3.3\left( {{{1.46}^x}} \right)} \right)\left( {{{0.62}^x}} \right)' - \left( {{{0.62}^x}} \right)\left( {6.42 + 3.3\left( {{{1.46}^x}} \right)} \right)'}}{{{{\left( {6.42 + 3.3\left( {{{1.46}^x}} \right)} \right)}^2}}}} \right) \cr
& {\text{compute derivatives}} \cr
& f'\left( x \right) = 430\left( {\frac{{\left( {6.42 + 3.3\left( {{{1.46}^x}} \right)} \right)\left( {{{0.62}^x}\ln 0.62} \right) - \left( {{{0.62}^x}} \right)\left( {3.3\left( {{{1.46}^x}} \right)\ln 1.46} \right)}}{{{{\left( {6.42 + 3.3\left( {{{1.46}^x}} \right)} \right)}^2}}}} \right) \cr
& {\text{factor out }}{0.62^x} \cr
& f'\left( x \right) = 430\left( {{{0.62}^x}} \right)\left( {\frac{{\left( {6.42 + 3.3\left( {{{1.46}^x}} \right)} \right)\left( {\ln 0.62} \right) - \left( {3.3\left( {{{1.46}^x}} \right)\ln 1.46} \right)}}{{{{\left( {6.42 + 3.3\left( {{{1.46}^x}} \right)} \right)}^2}}}} \right) \cr} $$
