If y = f(u) and u = g(x),where f and g are twice differentiable functions,show thatd2y/dx2 =d2y/du2 (du/dx)＾2 +dy/du(d2u/dx2)f 和 g 两次微分
网友回答
use chain rule.
u=g(x) ==> dg/dx=du/dx=u'
y=f(u) ==> df/dx=dy/du*du/dx=dy/du*u'
d2y/dx2=d(df/dx)/dx=d(dy/du)/dx*u'+dy/du* du'/dx
=d(dy/du)/du*du/dx*u'+dy/du*d2u/dx2
=d2y/du2*du/dx*du/dx+dy/du*d2u/dx2
另外twice differentiable functions意思是“二次可导函数”