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# 有關Chain rule的問題，煩請各位幫忙!!!

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~~

### 1 Answer

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- myisland8132Lv 71 decade agoFavorite Answer
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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