請問二次微分元平方的問題

2014-11-25 11:57 pm
令D=(x)•(d/dX)請問為什麼
(D^2)y=(x^2) • [(d^2)y / d(x^2)] + x•(dy/dX)

求詳解怎麼算

回答 (3)

2014-11-26 1:27 am
✔ 最佳答案
Dy =x d/dx (y) = x dy/dx
(D^2)y = D(Dy) = x d/dx ( x dy/dx)
=x { d (xdy/dx) /dx }
= x{ (dx/dx)(dy/dx) +x d(dy/dx) /dx }
= x{ (1)(dy/dx) +x [ (d^2)y /d(x^2) ] }
=x(dy/dx) +(x^2) [ (d^2)y /d(x^2) ]




{ f(x)g(x) } ' = f ' (x) g(x) + f(x) g '(x)
2014-11-26 2:12 am
若D(x)在x=0處可導,則函數在x=0處可導 導數為2x•D(x)+x^2•D'(x)=0 若D(x)在x=0處不...

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2014-11-26 1:23 am
D=x*(d/dx) => Dy=x*(dy/dx)D^2y=D(Dy)=D[x*(dy/dx)]=x*d[x*(dy/dx)]/dx=x*[(dA/dx)*B+A*(dB/dx)]......A=x, B=dy/dx=x*[(dx/dx)*(dy/dx)+x*d(dy/dx)/dx]=x*(1*dy/dx+x*d^2y/dx^2)=x*[x*d^2y/dx^2+(dy/dx)]=x^2*(d^2y/dx^2)+x*(dy/dx)=版主答案


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