Question

谁能帮忙解释下这个导数问题？

为什么要加这个y’，我写的话导完是这样ey+x*ey





Answer 1

y=1+xe^y
y'
= x(e^y)' + e^y
=x( e^y . y') + e^y
(1-xe^y)y' = e^y
y' = e^y/(1-xe^y)
=e^y/(2-y)
y''
=[(2-y)(e^y)' - e^y .(2-y)' ]/(2-y)^2
=[(2-y)(e^y .y') - e^y .(-y') ]/(2-y)^2
=y' .e^y.[ (2-y) +1 ]/(2-y)^2
=y' .e^y. (3-y)/(2-y)^2

追问

感谢解答！但我有一个疑问这里我看不懂可以在请教一下吗？1怎么变为2的

追答

u= e^y
du/dx
=d/dx ( e^y)
= e^y . dy/dx
= e^y . y'
//

y'=x( e^y . y') + e^y
y'-x( e^y . y') = e^y
y'(1-xe^y) =e^y
y' =e^y/(1-xe^y)