Consider the given function #f(x)=((x^2-5x^-1)/((5x)(-.5x)))^3(7x^6+4x^6-5x^2)#
We can split it into two functions #g(x)=((x^2-5x^-1)/((5x)(-.5x)))^3# and #h(x)=7x^6+4x^6-5x^2=11x^6-5x^2#.
Note that using product rule #(df)/(dx)=g'(x)h(x)+g(x)h'(x)#n and we know that #h'(x)=(dh)/(dx)=66x^5-10x#, we need to find #g'(x)#.
For #g(x)# we can use chain rule, as #g(x)=(u(x))^3# and hence
#(dg)/(dx)=3(u(x))^2=3((x^2-5x^-1)/((5x)(-.5x)))^2xx(du)/(dx)#
and as #u(x)=(x^2-5x^-1)/((5x)(-.5x))=-(x^2-1/x)/(2.5x^2)#
Now for finding #(du)/(dx)#, we can either have #u=-1/2.5+1/(2.5x^3)=-0.4+0.4x^(-3)#, which gives us #(du)/(dx)=-1.2/x^4# or use quotient formula, as follows
#(du)/(dx)=-(2.5x^2(2x+1/x^2)-5x(x^2-1/x))/(6.25x^4)#
= #-(5x^3+2.5-5x^3+5)/(6.25x^4)#
= #-7.5/(6.25x^4)=-1.2/x^4#
Hence #(dg)/(dx)=3((x^2-5x^-1)/((5x)(-.5x)))^2xx-1.2/x^4#
and #(df)/(dx)=3((x^2-5x^-1)/((5x)(-.5x)))^2xx-1.2/x^4(11x^6-5x^2)+(66x^5-10x)((x^2-5x^-1)/((5x)(-.5x)))^3#
and then simplified as indicated above.
