有關Chain rule的問題,煩請各位幫忙!!!

2006-12-06 10:33 am
i have a question like this
d/dx[sin(1/(x^2+1)^6)]

let f=[sin(1/(x^2+1)^6)]
v=(1/(x^2+1)^6)

Chain rule equation 應該係咁樣 df/dx=df/dv .dv/dx
但係我有一個咁o既step

=df/dv .dv/d(x^2+1) .d(x^2+1)/dx

點解dv/dx個part要變做=>dv/d(x^2+1) .d(x^2+1)/dx
可唔可以解釋一下依part係咩黎同埋係咪一定要咁做架??
thx~~

回答 (1)

2006-12-06 11:26 am
✔ 最佳答案
i have a question like this
d/dx[sin(1/(x^2+1)^6)]

let f=[sin(1/(x^2+1)^6)]
v=(1/(x^2+1)^6)

Chain rule equation 應該係咁樣 df/dx=df/dv .dv/dx
但係我有一個咁o既step

=df/dv .dv/d(x^2+1) .d(x^2+1)/dx

點解dv/dx個part要變做=>dv/d(x^2+1) .d(x^2+1)/dx
可唔可以解釋一下依part係咩黎同埋係咪一定要咁做架??

因為v的樣子是=(1/(x^2+1)^6)
就0甘dv/dx做唔到
所以要令u=x^2+1
dv/dx變成dv/du*du/dx
其實都係Chain rule﹐不過總共要3次
df/dx=df/dv .dv/du.du/dx
一定要0甘做的



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