Differentiate:
y=[x+(x^2+1)^(1/2)]^(1/2)
點計??
Differentiation問題
2008-05-25 8:45 pm
回答 (3)
2008-05-25 9:25 pm
As follow AS~~
As follow AS~~
圖片參考：http://i302.photobucket.com/albums/nn113/ken123572003/question4.jpg
As follow AS~~
圖片參考：http://i302.photobucket.com/albums/nn113/ken123572003/question4.jpg
2008-05-25 9:07 pm
y=[x+(x^2+1)^(1/2)]^(1/2)
dy/dx = {1/2*[x+(x^2+1)^(1/2)]^(1/2)} [1+2x/2sqrt(x^2+1)]
dy/dx = {1/2*[x+(x^2+1)^(1/2)]^(1/2)} [1+x/sqrt(x^2+1)]
2008-05-25 13:10:01 補充：
Chain Rule: dy/dx = dy/dz * dz/dx
let z = x+(x^2+1)^(1/2)
so,
dy/dx = 1/2*[x+(x^2+1)^(1/2)]^(1/2)
dz/dx = 1+[x/sqrt(x^2+1)]
dy/dx = {1/2*[x+(x^2+1)^(1/2)]^(1/2)} [1+2x/2sqrt(x^2+1)]
dy/dx = {1/2*[x+(x^2+1)^(1/2)]^(1/2)} [1+x/sqrt(x^2+1)]
2008-05-25 13:10:01 補充：
Chain Rule: dy/dx = dy/dz * dz/dx
let z = x+(x^2+1)^(1/2)
so,
dy/dx = 1/2*[x+(x^2+1)^(1/2)]^(1/2)
dz/dx = 1+[x/sqrt(x^2+1)]
收錄日期: 2021-04-15 15:19:08
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080525000051KK01123