y = sinx y' = lim(Δx→0) [sin(x + Δx) - sinx]/Δx = lim(Δx→0) 2cos[(x + Δx + x)/2]sin[(x + Δx - x)/2]/Δx = lim(Δx→0) 2cos(x + Δx/2)sin(Δx/2)/Δx = lim(Δx→0) cos(x + Δx/2) · sin(Δx/2)/(Δx/2) = cos(x + 0) · 1 = cosx
本文地址: http://www.goggeous.com/i/1/804046
文章来源:天狐定制
版权声明:除非特别标注,否则均为本站原创文章,转载时请以链接形式注明文章出处。
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2024-12-22 00:36:57职业培训
2024-12-22 00:36:56职业培训
2024-12-22 00:36:56职业培训
2024-12-22 00:36:55职业培训
2024-12-22 00:36:47职业培训
2024-12-22 00:36:47职业培训
2024-12-22 00:36:47职业培训
2024-12-22 00:36:46职业培训
2024-12-22 00:36:45职业培训
2024-12-22 00:36:45职业培训
2024-12-27 14:00职业培训
2025-01-03 07:39职业培训
2024-12-06 00:35职业培训
2024-12-28 07:40职业培训
2025-01-01 21:49职业培训
2025-01-05 05:00职业培训
2024-11-27 01:23职业培训
2024-12-18 04:18职业培训
2024-12-23 03:23职业培训
2024-12-21 23:46职业培训
扫码二维码
获取最新动态