Answer
$$f'\left( x \right) = \frac{{{{\left( {x + 1} \right)}^{3/2}}{{\left( {x - 4} \right)}^{5/2}}}}{{{{\left( {5x + 3} \right)}^{2/3}}}}\left( {\frac{3}{{2\left( {x + 1} \right)}} + \frac{5}{{2\left( {x - 4} \right)}} - \frac{{10}}{{3\left( {5x + 3} \right)}}} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{{\left( {x + 1} \right)}^{3/2}}{{\left( {x - 4} \right)}^{5/2}}}}{{{{\left( {5x + 3} \right)}^{2/3}}}} \cr
& {\text{taking the natural logarithm of both sides of the equation}} \cr
& \ln \left( {f\left( x \right)} \right) = \ln \left( {\frac{{{{\left( {x + 1} \right)}^{3/2}}{{\left( {x - 4} \right)}^{5/2}}}}{{{{\left( {5x + 3} \right)}^{2/3}}}}} \right) \cr
& {\text{quotient rule for logarithms}} \cr
& \ln \left( {f\left( x \right)} \right) = \ln \left( {{{\left( {x + 1} \right)}^{3/2}}{{\left( {x - 4} \right)}^{5/2}}} \right) - \ln \left( {{{\left( {5x + 3} \right)}^{2/3}}} \right) \cr
& {\text{product rule for logarithms}} \cr
& \ln \left( {f\left( x \right)} \right) = \ln \left( {{{\left( {x + 1} \right)}^{3/2}}} \right) + \ln \left( {{{\left( {x - 4} \right)}^{5/2}}} \right) - \ln \left( {{{\left( {5x + 3} \right)}^{2/3}}} \right) \cr
& {\text{power property for logarithms}} \cr
& \ln \left( {f\left( x \right)} \right) = \frac{3}{2}\ln \left( {x + 1} \right) + \frac{5}{2}\ln \left( {x - 4} \right) - \frac{2}{3}\ln \left( {5x + 3} \right) \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = \frac{3}{2}\left( {\frac{1}{{x + 1}}} \right) + \frac{5}{2}\left( {\frac{1}{{x - 4}}} \right) - \frac{2}{3}\left( {\frac{5}{{5x + 3}}} \right) \cr
& {\text{simplifying}} \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = \frac{3}{{2\left( {x + 1} \right)}} + \frac{5}{{2\left( {x - 4} \right)}} - \frac{{10}}{{3\left( {5x + 3} \right)}} \cr
& {\text{solving for }}f'\left( x \right) \cr
& f'\left( x \right) = f\left( x \right)\left( {\frac{3}{{2\left( {x + 1} \right)}} + \frac{5}{{2\left( {x - 4} \right)}} - \frac{{10}}{{3\left( {5x + 3} \right)}}} \right) \cr
& {\text{replace }}f\left( x \right){\text{ with the original function:}} \cr
& f'\left( x \right) = \frac{{{{\left( {x + 1} \right)}^{3/2}}{{\left( {x - 4} \right)}^{5/2}}}}{{{{\left( {5x + 3} \right)}^{2/3}}}}\left( {\frac{3}{{2\left( {x + 1} \right)}} + \frac{5}{{2\left( {x - 4} \right)}} - \frac{{10}}{{3\left( {5x + 3} \right)}}} \right) \cr} $$
