回覆列表
  • 1 # 已經過期了

    y=x^(2/3),根據導數基本定義,f"(x)=lim(h→0) [f(x+h)-f(x)]/h

    導數y"=lim(h→0) [(x+h)^(2/3)-x^(2/3)]/h

    分子:[(x+h)^(1/3)+x^(1/3)]*[(x+h)^(1/3)-x^(1/3)],這裡是平方差公式

    分母:h

    =lim(h→0)

    分子:[(x+h)^(1/3)+x^(1/3)][(x+h)^(2/3)-(x+h)^(1/3)*x^(1/3)+x^(2/3)]*[(x+h)^(1/3)-x^(1/3)][(x+h)^(2/3)+(x+h)^(1/3)*x^(1/3)+x^(2/3)],

    分母:h*[(x+h)^(2/3)+(x+h)^(1/3)*x^(1/3)+x^(2/3)]*[(x+h)^(2/3)-(x+h)^(1/3)*x^(1/3)+x^(2/3)]

    =lim(h→0)

    分子:[(x+h)+x)][(x+h)-x]

    分母:h*[(x+h)^(2/3)+(x+h)^(1/3)*x^(1/3)+x^(2/3)]*[(x+h)^(2/3)-(x+h)^(1/3)*x^(1/3)+x^(2/3)]

    =lim(h→0)

    分子:2x+h

    分母:[(x+h)^(2/3)+(x+h)^(1/3)*x^(1/3)+x^(2/3)]*[(x+h)^(2/3)-(x+h)^(1/3)*x^(1/3)+x^(2/3)],約去h

    =1/[x^(2/3)+x^(2/3)+x^(2/3)][x^(2/3)-x^(2/3)+x^(2/3)]*(2x)

    =1/[3x^(2/3)*x^(2/3)]*2x

    =2/3*x/x^(4/3)

    =2/3*1/x^(1/3)

    =2/[3x^(1/3)]

    因此y=x^(2/3)的導數為2/3*x^(-1/3)

  • 中秋節和大豐收的關聯?
  • 那些停留在你嘴裡和筆下的“別人”,是另一個無法釋然的自己嗎?