∇f/g = g∇f-f∇g / g2 係如何證的？
回答 (3)
2012-05-11 5:37 pm
✔ 最佳答案
Using the definition of ∇operator , 圖片參考：http://i286.photobucket.com/albums/ll87/hermittion/vectoridentityproof.jpg
2012-05-11 09:39:55 補充：
第一位回答者搞錯了!
Copestone 兄的提示是正確的
最近很忙所以剛剛才打字完成證明!
The proof may helpful !
2012-05-05 5:28 am
我假設你談的是 R^n 上的實函數。
∇其實是一個 vector field，(∂/∂x_1, ..., ∂/∂x_n);
∇f = (∂f/∂x_1, ..., ∂f/∂x_n);
可利用一維的微分法則。
∇(f/g) = (∂(f/g)/∂x_1, ...,∂(f/g)/∂x_n)
= (1/g^2) (g∂f/∂x_1 - f∂g/∂x_1, ..., g∂f/∂x_n - f∂g/∂x_n)
= (1/g^2) (g∇f - f∇g).
∇其實是一個 vector field，(∂/∂x_1, ..., ∂/∂x_n);
∇f = (∂f/∂x_1, ..., ∂f/∂x_n);
可利用一維的微分法則。
∇(f/g) = (∂(f/g)/∂x_1, ...,∂(f/g)/∂x_n)
= (1/g^2) (g∂f/∂x_1 - f∂g/∂x_1, ..., g∂f/∂x_n - f∂g/∂x_n)
= (1/g^2) (g∇f - f∇g).
2012-05-05 5:01 am
我見你問的式像是微分的商法則 若有誤會請見諒:
d(f/g) /dx=d(fg^-1)/dx
=f d(g^-1)/dx + g^-1 df/dx
=f d(g^-1)/dg *dg/dx +g^-1 df/dx
=f*-g^-2 *dg/dx + g^-1 df/dx
=[(-f dg)/dx]*g^-2+(g^-1*g^2 df/dx)*g^-2
=[-f dg/dx + g df/dx] / g^2
=[g df/dx - f dg/dx] / g^2
我想這證明應該幫到你吧
d(f/g) /dx=d(fg^-1)/dx
=f d(g^-1)/dx + g^-1 df/dx
=f d(g^-1)/dg *dg/dx +g^-1 df/dx
=f*-g^-2 *dg/dx + g^-1 df/dx
=[(-f dg)/dx]*g^-2+(g^-1*g^2 df/dx)*g^-2
=[-f dg/dx + g df/dx] / g^2
=[g df/dx - f dg/dx] / g^2
我想這證明應該幫到你吧
收錄日期: 2021-04-13 18:40:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120504000051KK00510