Ling asked in 科學及數學數學 · 1 decade ago

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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  • 1 decade ago
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    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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