Derivative

By first principle:

d(tan-1 x)/dx = lim (h → 0) [tan-1 (x + h) - tan-1 x]/h

Let θ = tan-1 (x + h) - tan-1 x, then

tan θ = [(x + h) - x]/[1 + (x + h)x] = h/[1 + x(x + h)]

θ = tan-1 {h/[1 + x(x + h)]}

Hence

d(tan-1 x)/dx = lim (h → 0) tan-1 {h/[1 + x(x + h)]}/h

= lim (h → 0) {{1/[1 + x(x + h)]} tan-1 {h/[1 + x(x + h)]}/{h/[1 + x(x + h)]}}

= lim (h → 0) 1/[1 + x(x + h)] lim (h → 0) tan-1 {h/[1 + x(x + h)]}/{h/[1 + x(x + h)]}

= 1/(1 + x2) lim (h → 0) tan-1 {h/[1 + x(x + h)]}/{h/[1 + x(x + h)]}

Sub u = tan-1 {h/[1 + x(x + h)]}, then u → 0 as h → 0

lim (h → 0) tan-1 {h/[1 + x(x + h)]}/{h/[1 + x(x + h)]} = lim (h → 0) u/tan u

= lim (h → 0) (u cos u/sin u)

= 1

Finally d(tan-1 x)/dx = 1/(1 + x2)

2012-06-22 16:27:40 補充：
Or can refer to the pic below:
http://i1191.photobucket.com/albums/z467/robert1973/Jun12/Crazydiff3.jpg