dx/1-2cosx+a^2?
回答 (1)
2018-07-09 12:41 pm
(d/dx)[f(x) / g(x)] = [g(x) f'(x) - f(x) g'(x)] / [g(x)]²
(d/dx)[x / (1 - 2cosx + a²)]
= [(1 - 2 cosx + a²) * (d/dx)x + x * (d/dx)(1 - 2 cosx + a²)] / (1 - 2 cosx + a²)²
= [(1 - 2 cosx + a²) * 1 + x * (-2 sinx)] / (1 - 2 cosx + a²)²
= (1 - 2 cosx - 2x sinx + a²) / (1 - 2cosx + a²)²
(d/dx)[x / (1 - 2cosx + a²)]
= [(1 - 2 cosx + a²) * (d/dx)x + x * (d/dx)(1 - 2 cosx + a²)] / (1 - 2 cosx + a²)²
= [(1 - 2 cosx + a²) * 1 + x * (-2 sinx)] / (1 - 2 cosx + a²)²
= (1 - 2 cosx - 2x sinx + a²) / (1 - 2cosx + a²)²
收錄日期: 2021-05-01 22:26:24
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180709043022AAvWqoP