大一高数下多元函数微分证明问题!难f(x,y)在D内对x连续,对(x,y'),(x,y'')有|[f(x,y')-f(x,y'')]/(y'-y'')|
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/23 17:18:29
![大一高数下多元函数微分证明问题!难f(x,y)在D内对x连续,对(x,y'),(x,y'')有|[f(x,y')-f(x,y'')]/(y'-y'')|](/uploads/image/z/2732339-11-9.jpg?t=%E5%A4%A7%E4%B8%80%E9%AB%98%E6%95%B0%E4%B8%8B%E5%A4%9A%E5%85%83%E5%87%BD%E6%95%B0%E5%BE%AE%E5%88%86%E8%AF%81%E6%98%8E%E9%97%AE%E9%A2%98%21%E9%9A%BEf%28x%2Cy%29%E5%9C%A8D%E5%86%85%E5%AF%B9x%E8%BF%9E%E7%BB%AD%2C%E5%AF%B9%28x%2Cy%27%29%2C%28x%2Cy%27%27%29%E6%9C%89%7C%5Bf%28x%2Cy%27%29-f%28x%2Cy%27%27%29%5D%2F%28y%27-y%27%27%29%7C)
大一高数下多元函数微分证明问题!难f(x,y)在D内对x连续,对(x,y'),(x,y'')有|[f(x,y')-f(x,y'')]/(y'-y'')|
大一高数下多元函数微分证明问题!难
f(x,y)在D内对x连续,对(x,y'),(x,y'')有|[f(x,y')-f(x,y'')]/(y'-y'')|
大一高数下多元函数微分证明问题!难f(x,y)在D内对x连续,对(x,y'),(x,y'')有|[f(x,y')-f(x,y'')]/(y'-y'')|
请看图片!
很简单!
f(x',y')-f(x'',y'')=[f(x',y')-f(x',y'')]+[f(x',y'')-f(x'',y'')]
分开用连续的定义来做!
对任意c>0 要使|f(x,y)-f(x0,y0)|
=|f(x,y)-f(x,y0)+f(x,y0)-f(x0,y0)|
<=|f(x,y)-f(x,y0)|+|f(x,y0)-f(x0,y0)|
<=L|y-y0|+|f(x,y0)-f(x0,y0)|
只要L|y-y0|
全部展开
对任意c>0 要使|f(x,y)-f(x0,y0)|
=|f(x,y)-f(x,y0)+f(x,y0)-f(x0,y0)|
<=|f(x,y)-f(x,y0)|+|f(x,y0)-f(x0,y0)|
<=L|y-y0|+|f(x,y0)-f(x0,y0)|
只要L|y-y0|
由于(x0,y0)的任意性可知函数在D上连续
收起