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求函数y=sinxcosx的单调区间求y=sinxcosx的单调

2018-05-30 02:55:27z***
求y=sinx cosx的单调区间用导数的方法~~求函数y=sinxcosx的单调区间求y=sinxcosx的单调区间用导数的方法~~:y=sinx cosxy'=cosx-sinx=√2[√2/2c?

最佳回答

  • y=sinx cosx
    y'=cosx-sinx=√2[√2/2cosx-√2/2sinx]=√2cos(x π/4)
    下面求单调递增区间
    y'>=0,
    2kπ-π/2 ≤x π/4 ≤ 2kπ π/2
    2kπ-3π/4 ≤x≤ 2kπ π/4
    单调递增区间[2kπ-3π/4 , 2kπ π/4]
    下面求单调递减区间
    y'<=0
    2kπ π/2 ≤x π/4 ≤ 2kπ 3π/2
    2kπ π/4 ≤x≤ 2kπ 4π/3
    单调递减区间[2kπ π/4 , 2kπ 4π/3]
    2018-05-30 04:21:12
  • 很赞哦! (77)