高阶导数问题y = f(x)的反函数为x = g(y)若g'(y) = 1/f'(x),求g'''(y)不太明白:g''(y)=(1/f'(x))'*(1/f'(x))后面为什么要多个*(1/f'(x))?
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高阶导数问题y = f(x)的反函数为x = g(y)若g'(y) = 1/f'(x),求g'''(y)不太明白:g''(y)=(1/f'(x))'*(1/f'(x))后面为什么要多个*(1/f'(x))?
高阶导数问题
y = f(x)的反函数为x = g(y)
若g'(y) = 1/f'(x),求g'''(y)
不太明白:g''(y)=(1/f'(x))'*(1/f'(x))
后面为什么要多个*(1/f'(x))?
高阶导数问题y = f(x)的反函数为x = g(y)若g'(y) = 1/f'(x),求g'''(y)不太明白:g''(y)=(1/f'(x))'*(1/f'(x))后面为什么要多个*(1/f'(x))?
得到g''(y)=(1/f'(x))'*(1/f'(x))=-f''(x)/(f'(x))^3
再对二阶导数求导:
g'''(x)=[-y'''y'+3(y'')^2]/(y')^5
多了这个是因为反函数是把y看成自变量,但f'(x)还是x的表达式,所以就还要求一次导数.就相当于复合函数求导了.如果这还不明白,可以用微分形式的证明.例如反函数二阶导数可以这样做:(dx)^2/dy^2=d(dx/dy)/dy=(d(dx/dy)/dx)*(dx/dy),而这后面的dx/dy=1/f'(x).可能我写的比较乱,你在草稿纸上好好写写,相信你能看得懂
x = g(y),两边对x求导,y看成x的函数用锁链法则
1=g'(y)y',
再求导有0=g''(y)(y')^2+g'(y)y'',
再求导有0=g'''(y)(y')^3+g''(y)*2y'y''+g''(y)y'y''+g'(y)y'''
消去g'(y)和g''(y)有
g'''(y)=3(y'')^2/(y')^5-y'''/(y')^4