两边取对数得： lnx+f(y)=y
两边对x求导：1/x+y'f'(y)=y',
得：y'=1/[x(1-f'(y))]
再对x求导：y"=-1/[x(1-f'(y))]^2* [1-f'(y)-xy'f"(y)]
代入y'得：y"=-1/[x(1-f'(y))]^2* [1-f'(y)-f"(y)/(1-f'(y))]
=[f"(y)-(1-f'(y))^2]/[x^2(1-f'(y))^3]
【隐函数求导】各位老师和高数大神,你们好!这道题,希望能帮助讲解.谢谢!十分感谢!
两边取对数得： lnx+f(y)=y
两边对x求导：1/x+y'f'(y)=y',
得：y'=1/[x(1-f'(y))]
再对x求导：y"=-1/[x(1-f'(y))]^2* [1-f'(y)-xy'f"(y)]
代入y'得：y"=-1/[x(1-f'(y))]^2* [1-f'(y)-f"(y)/(1-f'(y))]
=[f"(y)-(1-f'(y))^2]/[x^2(1-f'(y))^3]