Answer
$$f'\left( x \right) = \frac{{{x^8}{{\cos }^3}x}}{{\sqrt {x - 1} }}\left( {\frac{8}{x} - 3\tan x - \frac{1}{{2\left( {x - 1} \right)}}} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{x^8}{{\cos }^3}x}}{{\sqrt {x - 1} }} \cr
& {\text{rewrite the denominator}} \cr
& f\left( x \right) = \frac{{{x^8}{{\cos }^3}x}}{{{{\left( {x - 1} \right)}^{1/2}}}} \cr
& {\text{taking the natural logarithm of both sides of the equation}} \cr
& \ln \left( {f\left( x \right)} \right) = \ln \left( {\frac{{{x^8}{{\cos }^3}x}}{{{{\left( {x - 1} \right)}^{1/2}}}}} \right) \cr
& {\text{quotient rule for logarithms}} \cr
& \ln \left( {f\left( x \right)} \right) = \ln \left( {{x^8}{{\cos }^3}x} \right) - \ln {\left( {x - 1} \right)^{1/2}} \cr
& {\text{product rule for logarithms}} \cr
& \ln \left( {f\left( x \right)} \right) = \ln \left( {{x^8}} \right) + \ln \left( {{{\cos }^3}x} \right) - \ln {\left( {x - 1} \right)^{1/2}} \cr
& {\text{power property for logarithms}} \cr
& \ln \left( {f\left( x \right)} \right) = 8\ln \left( x \right) + 3\ln \left( {\cos x} \right) - \frac{1}{2}\ln \left( {x - 1} \right) \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = 8\left( {\frac{1}{x}} \right) + 3\left( {\frac{{ - \sin x}}{{\cos x}}} \right) - \frac{1}{2}\left( {\frac{1}{{x - 1}}} \right) \cr
& {\text{simplifying}} \cr
& \frac{{f'\left( x \right)}}{{f\left( x \right)}} = \frac{8}{x} - 3\tan x - \frac{1}{{2\left( {x - 1} \right)}} \cr
& {\text{solving for }}f'\left( x \right) \cr
& f'\left( x \right) = f\left( x \right)\left( {\frac{8}{x} - 3\tan x - \frac{1}{{2\left( {x - 1} \right)}}} \right) \cr
& {\text{replace }}f\left( x \right){\text{ with the original function:}} \cr
& f'\left( x \right) = \frac{{{x^8}{{\cos }^3}x}}{{\sqrt {x - 1} }}\left( {\frac{8}{x} - 3\tan x - \frac{1}{{2\left( {x - 1} \right)}}} \right) \cr} $$
