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求导数y=ln根号(x+1)/(x-1)

2010-11-13 14:49:37a***
求导数y=ln根号(x+1)/(x-1)求导数y=ln根号(x+1)/(x-1)：你的问题....？是求导数y=[ln(x+1)^(1/2)]/(x-1)还是求导数y=ln[(x+1)/(x-1)]^?

最佳回答

  • 你的问题....？是求导数y=[ln(x+1)^(1/2)]/(x-1)还是求导数y=ln[(x+1)/(x-1)]^(1/2) 求导数y=[ln(x+1)^(1/2)]/(x-1) y’={[(x-1)/2(x+1)]-[ln(x+1)^(1/2)]}/(x-1)^2 求导数y=ln[(x+1)/(x-1)]^(1/2) y’={1/[(x+1)/(x-1)]^(1/2)}*(1/2)[(x+1)/(x-1)]^(-1/2)*[x-1-(x+1)]/(x-1)^2 =[(1/2)*(x-1)*(-2)]/[(x+1)*(x-1)^2] =1/(1-x^2)
    2010-11-13 15:13:47
  • y=ln√[(x+1)/(x-1)] =(1/2)ln[(x+1)/(x-1)] =[ln(x+1)-ln(x-1)]/2 ∴y'=[1/(x+1)-1/(x-1)]/2 =[(x-1)-(x+1)]/[(x+1)(x-1)]/2 ={-2/[(x+1)(x-1]]}/2 =1/(1-x^2)
    2010-11-13 15:14:59
    • 很赞哦！ (26)

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