Question

试从dx/dy=1/y'中导出d2x/dy2=-y''/(y')3

为什么画圈的那部分说要用到复合函数的求导？

Answer 1

dx/dy = 1/y'
d^2x /dy^2
= d/dy ( dx/dy)
= d/dy (1/y')
=[d/dx ( 1/y') ] . dx/dy ( 链式法则)
=[d/dx ( 1/y') ] . ( 1/y')
= [-y''/(y')^2] . ( 1/y')
=-y''/(y')^3

追问

怎么从倒数第三步推出倒二的呀

追答

d/dx ( 1/y')
=d/dx ( y')^(-1)
= -(y')^(-2) . d/dx (y')
= -(y')^(-2) . y''
=-y''/(y')^2

追问

是因为倒四的时候把式子变成了复合函数所以这里要用复合函数求导吗

追答

y' = dy/dx
y'' = d^2y/dx^2
d/dx ( 1/y')
=d/dx [ (dy/dx)^(-1) ]
= (-1). (dy/dx)^(-2) . d/dx [(dy/dx)] ( 链式法则)
=- (dy/dx)^(-2) . (d^2y/dx^2 )
=-(y')^(-2) . y''
=-y''/(y')^2

追问

明白啦谢谢啦