设f(x)为单调二阶可导函数,其反函数为g(x),且已知f(1)=2f'(1)=-3^(1/2)f''(1)=1求g''(2)

Question

设f(x)为单调二阶可导函数,其反函数为g(x),且已知f(1)=2 f'(1)=-3^(1/2) f''(1)=1 求g''(2)

Answer 1

g[f(x)]=x
g(2)=1
{g[f(x)]}'=f'(x)*g'[f(x)]=1
{g[f(x)]}''=f''(x)*g'[f(x)]+[f'(x)]^2*g''[f(x)]=0
带入x=1
f'(1)*g'(2)=1 g'(2)=1/f'(1)
f''(1)*g'(2)+[f'(1)]^2*g''(2)=0
g"(2)=-f''(1)*g'(2)/[f'(1)]^2=-g'(2)/[f'(1)]^2=-1/[f'(1)]^3=3^(-3/2)
不知对否 楼主看下