Lawliet
DOTA2游廊地图作者
 
 
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Spxctre Jan 22, 2023 @ 2:56am 
♥♥♥♥♥♥ те в ачько. :empirecross:
Дамский угодник Jan 21, 2023 @ 1:45am 
Неплох :bombsmile:
месси Jan 21, 2023 @ 12:04am 
ПРИВЕТ
xcxcxcsccxxcxxxccxcxcxxc Oct 2, 2022 @ 11:21am 
ало черт,твоя кастомка не фурычит,если ты ее не починишь,я найду тебя и трахну:steamhappy:
Lawliet May 4, 2022 @ 8:16am 
(sinx)' = cosx
(cosx)' = - sinx
(tanx)'=1/(cosx)^2=(secx)^2=1+(tanx)^2
-(cotx)'=1/(sinx)^2=(cscx)^2=1+(cotx)^2
(secx)'=tanx·secx
(cscx)'=-cotx·cscx
(arcsinx)'=1/(1-x^2)^1/2
(arccosx)'=-1/(1-x^2)^1/2
(arctanx)'=1/(1+x^2)
(arccotx)'=-1/(1+x^2)
(arcsecx)'=1/(|x|(x^2-1)^1/2)
(arccscx)'=-1/(|x|(x^2-1)^1/2)
④(sinhx)'=coshx
(coshx)'=sinhx
(tanhx)'=1/(coshx)^2=(sechx)^2
(coth)'=-1/(sinhx)^2=-(cschx)^2
(sechx)'=-tanhx·sechx
(cschx)'=-cothx·cschx
(arsinhx)'=1/(x^2+1)^1/2
(arcoshx)'=1/(x^2-1)^1/2
(artanhx)'=1/(x^2-1) (|x|<1)
(arcothx)'=1/(x^2-1) (|x|>1)
(arsechx)'=1/(x(1-x^2)^1/2)
(arcschx)'=1/(x(1+x^2)^1/2)