作者LuisSantos (^______^)
看板trans_math
标题Re: [微分] 93中正企管
时间Mon Dec 3 00:13:52 2007
※ 引述《bitchdog (PEaCE)》之铭言：
: y=[x^3+(5x+1)^6]^4 , find y'
y = (x^3 + (5x + 1)^6)^4
dy
y' = ----
dx
d
= (4)((x^3 + (5x + 1)^6)^3)(----(x^3 + (5x + 1)^6))
dx
= (4)((x^3 + (5x + 1)^6)^3)((3)(x^2) + (6)((5x + 1)^5)(5))
= (4)((x^3 + (5x + 1)^6)^3)((3)(x^2) + (30)((5x + 1)^5))
--
※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 61.66.173.21