y=sin(x+y),一階隱導數(shù)y'=cos(x+y)/[1-cos(x+y)]我懂,但如何求二階隱導數(shù)~
y=sin(x+y),一階隱導數(shù)y'=cos(x+y)/[1-cos(x+y)]我懂,但如何求二階隱導數(shù)~
sin(x+y)/[cos(x+y)-1]^3
sin(x+y)/[cos(x+y)-1]^3
數(shù)學人氣：396 ℃時間：2020-08-05 06:57:16
優(yōu)質解答
一階導數(shù)你求對了令 t = dy/dx = cos(x+y)/[1-cos(x+y)] dy = cos(x+y)·dx/[1-cos(x+y)] dx+dy = cos(x+y)·dx/[1-cos(x+y)] + dx = dx/[1-cos(x+y)] t+1 = 1+ cos(x+y)/[1-cos(x+y)] = 1/[1-cos(x+y)] 那么原函數(shù)...
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