※ 引述《kusorz (^~^)》之銘言：
: 設F(x)=tan^-1(x-1/x+1)-tan^-1x
: 試求F的導函數
[(x-1)/(x+1)]' 1
F'(x)=------------------------ - -----
1 + [(x-1)/(x+1)]^2 1+x^2
[(x+1)-(x-1)] / (x+1)^2 1
=------------------------------ - -------
1 + [(x-1)/(x+1)]^2 1+x^2
2 1
= ------------------ - -------
(x+1)^2 + (x-1)^2 1+x^2
2(1+x^2)-[(x+1)^2 + (x-1)^2]
= --------------------------------- = 0
[(x+1)^2 + (x-1)^2](1+x^2)
∵2(1+x^2)-[(x+1)^2 + (x-1)^2] = 2+2x^2-(2+2x^2)= 0
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 218.167.75.28