设y=f(x)与x=g（x）互为反函数,f'(x)=x^4+x^2+1那么g''(x)为多少?

Question

设y=f(x)与x=g（x）互为反函数,f'(x)=x^4+x^2+1那么g''(x)为多少?

Answer 1

反函数具有自反性,所以x=g(y)
两边求导：1＝（dg(y)/dx）=(dg(y)/dy)*(dy/dx)
化简：1/(x^4+x^2+1)=g'(y)
再导：d(1/(x^4+x^2+1))/dx=d(g'(y))/dx=d(g'(y))/dy*dy/dx
-(4x^3+2x)/(x^4+x^2+1)^2=g''(y)*(x^4+x^2+1)
化简g''(y)=-(4x^3+2x)/(x^4+x^2+1)^3
又因x=g(x)
所以g''(x)=-(4x^3+2x)/(x^4+x^2+1)^3