Question

y=sin(x+y)的微分怎么求？

Answer 1

该题型为隐函数微分问题，可以运用导数的链式法则和微分运算法则来求解。
解：
dy=d(sin(x+y))
=cos(x+y)·d(x+y)
=cos(x+y)·(dx+dy)
=cos(x+y)dx+cos(x+y)dy
dy-cos(x+y)dy=cos(x+y)dx
dy=cos(x+y)/(1-cos(x+y))dx
所以，
y=sin(x+y)的微分为 dy=cos(x+y)/(1-cos(x+y))dx

Answer 2

y= sin(x+y)
y'= ( 1+ y')cos(x+y)
y''=y''.cos(x+y) -(1+y')^2 .sin(x+y)
=y''.cos(x+y) -(1+y').y'
=y''.cos(x+y) -{ 1+ cos(x+y)/(1-cos(x+y) ] } .[cos(x+y)/[1-cos(x+y)]
=y''.cos(x+y) -{ cos(x+y)/[(1-cos(x+y) ]^2 }
[1-cos(x+y) ] y''=-cos(x+y)/[(1-cos(x+y) ]^2
y''=- cos(x+y)/[(1-cos(x+y) ]^3