ln 微分運算

d/dx[ln(x+√(1+x^2)]
=1/[x+√(1+x^2)]*(1+x/√(1+x^2))
=1/√(1+x^2)

2010-04-06 22:35:01 補充：
sorry for forgetting the square
because I can't see in the picture

{d/dx[ln(x+√(1+x^2)]}^2
={1/[x+√(1+x^2)]*(1+x/√(1+x^2))}^2
=[1/√(1+x^2)]^2
=1/(1+x^2)

http://i841.photobucket.com/albums/zz333/tammie_0666/2010-4-6-2.jpg?t=1270560787


圖片參考：http://i841.photobucket.com/albums/zz333/tammie_0666/2010-4-6-2.jpg?t=1270560787

for 002 answer...how about the square outside d/dx??

The steps are as follow:

圖片參考：http://imgcld.yimg.com/8/n/HA00238614/o/701004060153613873381000.jpg