y=f（x^2+1）,且f'（x）=x^2,则dy／dx | x=-1=?

df(x)/dx = x^2,df(x^2+1)/d(x^2+1) = (x^2+1)^2,dy/dx = df(x^2+1)/dx = df(x^2+1)/d(x^2+1)*d(x^2+1)/dx = (x^2+1)^2* 2x,dy/dx|_{x=-1} = [2x(x^2+1)^2]|_{x=-1}=[2(-1)(1+1)^2]=-8