To differentiate #f(x)=[(sinx+tanx)/(sinxcosx)]^3#,
we use chain rule and quotient rule.
Let #f(x)=(g(x))^3#, where #g(x)=(sinx+tanx)/(sinxcosx)#
Now as #f(x)=(g(x))^3#, #(df)/(dg)=3(g(x))^2#
and usiing quotient rule #(dg)/(dx)=(sinxcosx xx (cosx+sec^2x)-
(sinx+tanx)(cos^2x-sin^2x))/(sin^2xcos^2x)#
= #(sinxcos^2x+sinxsecx-sinxcos^2x-tanxcos^2x+sin^3x+sin^2xtanx)/(sin^2xcos^2x)#
= #(sinxcos^2x+tanx-sinxcos^2x-sinxcosx+sin^3x+sin^2xtanx)/(sin^2xcos^2x)#
= #1/sinx+1/(sinxcos^3x)-1/sinx-1/(sinxcosx)+sinx/cos^2x+sinx/cos^3x#
= #1/(sinxcos^3x)-1/(sinxcosx)+sinx/cos^2x+sinx/cos^3x#
= #(1-cos^2x)/(sinxcos^3x)+(sinx(cosx+1))/cos^3x#
= #sinx/cos^3x+(sinx(cosx+1))/cos^3x#
= #(sinx(cosx+2))/cos^3x#
and #(df)/(dx)=(df)/(dg)xx(dg)/(dx)#
= #3((sinx+tanx)/(sinxcosx))^2(sinx(cosx+2))/cos^3x#
