1,f(x)=(πtan sec x)^6
f'(x)=[6(πtan sec x)^5]×[πsec^2(sec x)]×[secx tanx]
=6π^6(tan secx)^5×(sec secx)^2×secx×tanx
令g(x)=πtan sec x
f(x)=[g(x)]^6
由復合函數(shù)的求導規(guī)則
f'(x)=6[g(x)]^5×g'(x)
g'(x)=πtan'(sec x)×sec'x
={π[sec (sec x)]^2}×(secx tanx).
2,f(x)=arcsin(sinx+1/2)
f'(x)={1/√[1-(sinx+1/2)^2]}×cos x.
令g(x)=sin x+1/2
這里要注意f(x)的定義域,-1≤sin x+1/2≤1
即-1≤sin x≤1/2.
定義域為：[-π/2+2kπ,π/6+2kπ]∪[5π/6+2kπ,2π+2kπ].
f'(x)=1/√{1-[g(x)]^2}×g'(x)
g'(x)=cos x.第一個好像不對吧，答案是6*Pi^6*tan(x)^5*sec(x)^6*(1+tan(x)^2)+6*Pi^6*tan(x)^7*sec(x)^6第一個不對，題目看錯了，一開始就寫錯了，等等先f(x)=(πtanx sec x)^6同樣的令g(x)=πtanx sec xf'(x)=6[g(x)]^5*g'(x);g'(x)=π[sec(x)*tan'(x)+tan(x)*sec'(x)]=π[sec(x)]^3+πsec(x)[tan(x)]^2所以f'(x)=6π^6*tan(x)^5*sec(x)^5*{sec(x)^3+sec(x)tan(x)^2}因為：sec(x)^2=1+tan(x)^2帶入化解，是樓主提供的答案，最開始的回答以為f(x)=(πtan sec x)^6，不好意思。